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    NTT

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct fastio{
 8     static const int N=12;
 9     int b[N],B;
10     int ru(){
11         int x=0;
12         char c=getchar();
13         while(c<48)c=getchar();
14         while(c>47)x=x*10+c-48,c=getchar();
15         return x;
16     }
17     int ri(){
18         int x=0,o=1;
19         char c=getchar();
20         while(c<45)c=getchar();
21         if(c=='-')o=-1,c=getchar();
22         while(c>47)x=x*10+c-48,c=getchar();
23         return x*o;
24     }
25     void wu(int x){
26         if(!x){
27             putchar('0');
28             return;
29         }
30         B=0;
31         while(x)b[++B]=x%10,x/=10;
32         while(B)putchar(48+b[B--]);
33     }
34     void wi(int x){
35         if(x<0)putchar('-'),x=-x;
36         if(!x){
37             putchar('0');
38             return;
39         }
40         B=0;
41         while(x)b[++B]=x%10,x/=10;
42         while(B)putchar(48+b[B--]);
43     }
44 }F;
45 struct FFT{
46     static const int N=262222,M=998244353ll;
47     int n,m;
48     ll a[N],b[N],w[N],w2[N],r;
49     ll P(ll x,ll k){
50         ll ans=1;
51         while(k){
52             if(k&1)ans=ans*x%M;
53             x=x*x%M,k>>=1;
54         }
55         return ans;
56     }
57     void fft(ll *a,ll *w){
58         for(int i=0,j=0;i<n;++i){
59             if(i<j)swap(a[i],a[j]);
60             for(int k=n>>1;(j^=k)<k;k>>=1);
61         }
62         for(int i=2;i<=n;i<<=1){
63             for(int j=0;j<n;j+=i){
64                 for(int k=0;k<(i>>1);++k){
65                     int l=j+k,r=l+(i>>1);
66                     ll o=a[r]*w[n/i*k]%M;
67                     a[r]=(a[l]+M-o)%M,a[l]=(a[l]+o)%M;
68                 }
69             }
70         }
71     }
72     void mul(){
73         fft(a,w),fft(b,w);
74         for(int i=0;i<n;++i)a[i]=a[i]*b[i]%M*r%M;
75         fft(a,w2);
76     }
77     void init(){
78         n=F.ru(),m=F.ru();
79         for(int i=0;i<n;++i)a[i]=F.ru();
80         for(int i=0;i<m;++i)b[i]=F.ru();
81         m+=n-2,n=1;
82         while(n<=m)n<<=1;
83         r=P(n,M-2),w[0]=w2[0]=1,w[1]=w2[n-1]=P(3,(M-1)/n);
84         for(int i=2;i<n;++i)w[i]=w2[n-i]=w[i-1]*w[1]%M;
85     }
86     void out(){
87         for(int i=0;i<m;++i)F.wu((int)a[i]),putchar(' ');
88         F.wu((int)a[m]),putchar('
');
89     }
90 }_;
91 int main(){
92     _.init(),_.mul(),_.out();
93     return 0;
94 }
    View Code

    dinic

    #include<bits/stdc++.h>
#define ll   long long
#define pb   push_back
#define mp   make_pair
#define orz  1000000007
using namespace std;
int gi(){
    char c=getchar();
    int x=0;
    while(c<48)c=getchar();
    while(c>47)x=x*10+c-48,c=getchar();
    return x;
}
struct E{int x,y,z;};
vector<E> v[10005];
void add(int x,int y,int z){
    E X,Y;
    X.x=y,X.y=z,X.z=v[y].size();
    Y.x=x,Y.y=0,Y.z=v[x].size();
    v[x].pb(X),v[y].pb(Y);
}
int n,m,s,t,ans;
int l[10005],i[10005],q[10005],L,R;
int dfs(int x,int y){
    if(x==t) return y;
    int ret=0;
    for(;i[x]<v[x].size();++i[x]){
        E &e=v[x][i[x]];
        if(!e.y||l[e.x]!=l[x]+1) continue;
        int d=dfs(e.x,min(y,e.y));
        if(!d) continue;
        e.y-=d,v[e.x][e.z].y+=d,ret+=d,y-=d;
        if(!y) return ret;
    }
    l[x]=-1;
    return ret;
}
int main(){
    n=gi();m=gi();s=gi();t=gi();
    for(int i=1;i<=m;++i){
        int x,y,z;
        x=gi();y=gi();z=gi();
        add(x,y,z);
    }
    while(1){
        memset(i,0,sizeof(i));
        memset(l,-1,sizeof(l));
        l[s]=0,L=0,R=1,q[1]=s;
        while(L!=R){
            int x=q[++L];
            for(int i=0;i<v[x].size();++i){
                if(v[x][i].y&&l[v[x][i].x]==-1)l[v[x][i].x]=l[x]+1,q[++R]=v[x][i].x;
            }
        }
        if(l[t]==-1) break;
        ans+=dfs(s,orz);
    }
    printf("%d
",ans);
    return 0;
}
    View Code

    费用流(dijkstra)

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct E{int x,y,z,c;};
 8 struct F{
 9     vector<E> v[5005];
10     int n,s,t,ans,res,h[5005],d[5005],pv[5005],pe[5005];
11     void upd(int &x,int y){if(x>y)x=y;}
12     void init(){
13         for(int i=1;i<=n;++i)v[i].clear();
14     }
15     void add(int x,int y,int z,int c){
16         E X,Y;
17         X.x=y,X.y=z,X.z=v[y].size(),X.c=c;
18         Y.x=x,Y.y=0,Y.z=v[x].size(),Y.c=-c;
19         v[x].pb(X),v[y].pb(Y);
20     }
21     void flow(){
22         ans=res=0;
23         memset(h,0,sizeof(h));
24         while(1){
25             for(int i=1;i<=n;++i)d[i]=orz;
26             d[s]=0;
27             priority_queue<pair<int,int> >q;
28             q.push(mp(0,s));
29             while(!q.empty()){
30                 int dd=-q.top().first,k=q.top().second;
31                 q.pop();
32                 if(d[k]!=dd) continue;
33                 for(int i=0;i<v[k].size();++i){
34                     E &e=v[k][i];
35                     if(e.y){
36                         int u=e.x,D=dd+e.c+h[k]-h[u];
37                         if(d[u]>D)d[u]=D,pv[u]=k,pe[u]=i,q.push(mp(-D,u));
38                     }
39                 }
40             }
41             if(d[t]==orz) return;
42             for(int i=1;i<=n;++i)h[i]+=d[i];
43             int o=orz;
44             for(int i=t;i!=s;i=pv[i])upd(o,v[pv[i]][pe[i]].y);
45             ans+=o,res+=o*h[t];
46             for(int i=t;i!=s;i=pv[i]){
47                 E &e=v[pv[i]][pe[i]];
48                 e.y-=o,v[e.x][e.z].y+=o;
49             }
50         }
51     }
52 }f;
53 int m;
54 int r(){
55     int x=0;
56     char c=getchar();
57     while(c<48)c=getchar();
58     while(c>47)x=x*10+c-48,c=getchar();
59     return x;
60 }
61 int main(){
62     f.n=r(),m=r(),f.s=r(),f.t=r(); 
63     for(int i=1;i<=m;++i){
64         int x,y,z,c;
65         x=r(),y=r(),z=r(),c=r();
66         f.add(x,y,z,c);
67     }
68     f.flow();
69     printf("%d %d
",f.ans,f.res);
70     return 0;
71 }
    View Code

    树链剖分

      1 #include<bits/stdc++.h>
  2 #define ll   long long
  3 #define pb   push_back
  4 #define mp   make_pair
  5 #define orz  1000000007
  6 using namespace std;
  7 int n,m,r,f[100005][18],d[100005],a[100005],s[100005],S=1,F[100005],P[100005];
  8 vector<int> v[100005];
  9 ll p;
 10 struct bit{
 11     ll t[100005],t2[100005];
 12     void add(int l,int r,int x){
 13         int k=l;
 14         ll o=(1-l)*1ll*x;
 15         while(k<=n)t[k]=(t[k]+o)%p,k+=(k&-k);
 16         k=l,o=x;
 17         while(k<=n)t2[k]=(t2[k]+o)%p,k+=(k&-k);
 18         k=r+1,o=r*1ll*x;
 19         while(k<=n)t[k]=(t[k]+o)%p,k+=(k&-k);
 20         k=r+1,o=-x;
 21         while(k<=n)t2[k]=(t2[k]+o)%p,k+=(k&-k);
 22     }
 23     int get(int l,int r){
 24         ll ans1=0,ans2=0;
 25         int k=r;
 26         while(k)ans2+=t2[k],k-=(k&-k);
 27         ans2*=r;
 28         k=r;
 29         while(k)ans2+=t[k],k-=(k&-k);
 30         k=l-1;
 31         while(k)ans1+=t2[k],k-=(k&-k);
 32         ans1*=(l-1);
 33         k=l-1;
 34         while(k)ans1+=t[k],k-=(k&-k);
 35         return (int)(((ans2-ans1)%p+p)%p);
 36     }
 37 }t;
 38     
 39 void ope(int x,int y){
 40     d[x]=d[y]+1,f[x][0]=y,s[x]=1;
 41     for(int i=0;i<17;++i)f[x][i+1]=f[f[x][i]][i];
 42     for(int i=0;i<v[x].size();++i)if(v[x][i]!=y)ope(v[x][i],x),s[x]+=s[v[x][i]];
 43 }
 44 int lca(int x,int y){
 45     if(d[x]<d[y])swap(x,y);
 46     for(int i=17;i>=0;--i)if(d[f[x][i]]>=d[y])x=f[x][i];
 47     if(x==y) return x;
 48     for(int i=17;i>=0;--i)if(f[x][i]!=f[y][i])x=f[x][i],y=f[y][i];
 49     return f[x][0];
 50 }
 51 void dfs(int x,int y){
 52     t.add(F[x],F[x],a[x]);
 53     int son=0;
 54     for(int i=0;i<v[x].size();++i)if(v[x][i]!=y&&s[v[x][i]]>s[son])son=v[x][i];
 55     if(!son) return;
 56     F[son]=++S,P[son]=P[x],dfs(son,x);
 57     for(int i=0;i<v[x].size();++i)if(v[x][i]!=y&&v[x][i]!=son)F[v[x][i]]=++S,P[v[x][i]]=v[x][i],dfs(v[x][i],x);
 58 }
 59 void change(int l,int r,int x){
 60     if(d[l]>d[r]) return;
 61     if(d[P[r]]>=d[l])t.add(F[P[r]],F[r],x),change(l,f[P[r]][0],x);
 62     else t.add(F[l],F[r],x);
 63 }
 64 int answer(int l,int r){
 65     if(d[l]>d[r]) return 0;
 66     if(d[P[r]]>=d[l]) return (t.get(F[P[r]],F[r])+answer(l,f[P[r]][0]))%p;
 67     return t.get(F[l],F[r]);
 68 }
 69 int main(){
 70     memset(t.t,0,sizeof(t.t));
 71     memset(t.t2,0,sizeof(t.t2));
 72     scanf("%d%d%d%lld",&n,&m,&r,&p);
 73     for(int i=1;i<=n;++i)scanf("%d",a+i);
 74     for(int i=1;i<n;++i){
 75         int x,y;
 76         scanf("%d%d",&x,&y);
 77         v[x].pb(y),v[y].pb(x);
 78     }
 79     ope(r,0),F[r]=1,P[r]=r,dfs(r,0);
 80     for(int i=1;i<=m;++i){
 81         int T,x,y,z,k;
 82         scanf("%d",&T);
 83         if(T==1){
 84             scanf("%d%d%d",&x,&y,&k);
 85             z=lca(x,y),change(z,x,k),change(z,y,k),change(z,z,-k);
 86         }
 87         else if(T==2){
 88             scanf("%d%d",&x,&y);
 89             z=lca(x,y),printf("%d
",(int)((((answer(z,x)+answer(z,y))%p)-answer(z,z)+p)%p));
 90         }
 91         else if(T==3){
 92             scanf("%d%d",&x,&k);
 93             t.add(F[x],F[x]+s[x]-1,k);
 94         }
 95         else{
 96             scanf("%d",&x);
 97             printf("%d
",t.get(F[x],F[x]+s[x]-1));
 98         }
 99     }
100     return 0;
101 }
    View Code

    无旋treap

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct treap{
 8     int size,val,key;
 9     treap* ch[2];
10 };
11 treap *empty,*root;
12 void init(treap *t,int x){t->val=x,t->key=rand(),t->size=1,t->ch[0]=t->ch[1]=empty;}
13 treap *merge(treap *x,treap *y){
14     if(x==empty) return y;
15     if(y==empty) return x;
16     if(x->key<y->key){x->size+=y->size,x->ch[1]=merge(x->ch[1],y);return x;}
17     y->size+=x->size,y->ch[0]=merge(x,y->ch[0]);return y;
18 }
19 pair<treap*,treap*> split(treap *t,int k){
20     if(t==empty) return mp(t,t);
21     pair<treap*,treap*> ret;
22     if(t->ch[0]->size>=k)ret=split(t->ch[0],k),t->ch[0]=ret.second,t->size-=k,ret.second=t;
23     else ret=split(t->ch[1],k-t->ch[0]->size-1),t->ch[1]=ret.first,t->size=k,ret.first=t;
24     return ret;
25 }
26 int rank(treap *t,int x){
27     if(t==empty) return 0;
28     if(t->val>=x) return rank(t->ch[0],x);
29     return rank(t->ch[1],x)+t->ch[0]->size+1;
30 }
31 int get(treap *t,int k){
32     if(t->ch[0]->size>=k) return get(t->ch[0],k);
33     if(t->ch[0]->size+1==k) return t->val;
34     return get(t->ch[1],k-t->ch[0]->size-1);
35 }
36 void ins(int x){
37     treap *v=new treap;
38     init(v,x);
39     pair<treap*,treap*> p=split(root,rank(root,x));
40     root=merge(merge(p.first,v),p.second);
41 }
42 void del(int k){
43     pair<treap*,treap*> p=split(root,k-1),q=split(p.second,1);
44     root=merge(p.first,q.second);
45 }
46 int main(){
47     empty=new treap,root=new treap;
48     empty->size=0,empty->val=0,empty->key=0,empty->ch[0]=empty->ch[1]=empty;
49     init(root,orz);
50     int T;
51     scanf("%d",&T);
52     while(T--){
53         int t,x;
54         scanf("%d%d",&t,&x);
55         if(t==1)ins(x);
56         if(t==2)del(rank(root,x)+1);
57         if(t==3)printf("%d
",rank(root,x)+1);
58         if(t==4)printf("%d
",get(root,x));
59         if(t==5)printf("%d
",get(root,rank(root,x)));
60         if(t==6)printf("%d
",get(root,rank(root,x+1)+1));
61     }
62     return 0;
63 }
    View Code

    二逼平衡树(CDQ分治+整体二分)

      1 #include<bits/stdc++.h>
  2 #define ll   long long
  3 #define pb   push_back
  4 #define mp   make_pair
  5 #define orz  1000000007
  6 using namespace std;
  7 /*
  8   CDQ:
  9   get rank-     t=0
 10   get rank      t=1
 11   insert        t=2
 12   remove        t=3
 13                       l=pos              k=num
 14   go:
 15   find kth      t=1   l=left   r=right   k=kth
 16   insert        t=2
 17   remove        t=3
 18 */
 19 struct S{int t,l,r,k,i;}p[500005],q[500005],q2[500005];
 20 struct BIT{int t[100005];}T;
 21 int n,m,a[50005],b[50005],t[50005],l[50005],r[50005],k[50005],ans[50005],P;
 22 pair<int,int> Q[100005];
 23 int back[100005],O;
 24 void I(int t,int l,int r,int k,int i){p[++P].t=t,p[P].l=l,p[P].r=r,p[P].k=k,p[P].i=i;}
 25 void add(int x,int k){while(x<=n)T.t[x]+=k,x+=(x&-x);}
 26 int sum(int x){
 27     int ans=0;
 28     while(x)ans+=T.t[x],x-=(x&-x);
 29     return ans;
 30 }
 31 void cdq(int l,int r){
 32     if(l>=r) return;
 33     int m=(l+r)>>1,L=l,R=m+1;
 34     cdq(l,m),cdq(m+1,r);
 35     for(int i=l;i<=r;++i){
 36         if(L<=m&&(R>r||p[L].k<p[R].k)){
 37             q[i]=p[L++];
 38             if(q[i].t==2)add(q[i].l,1);
 39             else if(q[i].t==3)add(q[i].l,-1);
 40         }
 41         else{
 42             q[i]=p[R++];
 43             if(q[i].t==0)ans[q[i].i]-=sum(q[i].l);
 44             else if(q[i].t==1)ans[q[i].i]+=sum(q[i].l);
 45         }
 46     }
 47     for(int i=l;i<=m;++i){
 48         if(p[i].t==2)add(p[i].l,-1);
 49         else if(p[i].t==3)add(p[i].l,1);
 50     }
 51     for(int i=l;i<=r;++i)p[i]=q[i];
 52 }
 53 void go(int l,int r,int L,int R){
 54     if(l>r||L>R) return;
 55     if(l==r){
 56         for(int i=L;i<=R;++i)if(p[i].t==1)ans[p[i].i]=back[l];
 57         return;
 58     }
 59     int m=(l+r)>>1,v=0,v2=0;
 60     for(int i=L;i<=R;++i){
 61         if(p[i].t==2){
 62             if(p[i].k>m)q2[++v2]=p[i];
 63             else q[++v]=p[i],add(p[i].l,1);
 64         }
 65         else if(p[i].t==3){
 66             if(p[i].k>m)q2[++v2]=p[i];
 67             else q[++v]=p[i],add(p[i].l,-1);
 68         }
 69         else{
 70             int o=sum(p[i].r)-sum(p[i].l-1);
 71             if(o>=p[i].k)q[++v]=p[i];
 72             else p[i].k-=o,q2[++v2]=p[i];
 73         }
 74     }
 75     for(int i=L;i<=R;++i){
 76         if(p[i].t==2){
 77             if(p[i].k<=m)add(p[i].l,-1);
 78         }
 79         else if(p[i].t==3){
 80             if(p[i].k<=m)add(p[i].l,1);
 81         }
 82     }
 83     for(int i=1;i<=v;++i)p[L+i-1]=q[i];
 84     for(int i=1;i<=v2;++i)p[R-v2+i]=q2[i];
 85     go(l,m,L,L+v-1),go(m+1,r,L+v,R);
 86 }
 87 int main(){
 88     scanf("%d%d",&n,&m);
 89     for(int i=1;i<=n;++i)scanf("%d",a+i),Q[i]=mp(a[i],i);
 90     for(int i=1;i<=m;++i){
 91         scanf("%d",t+i);
 92         if(t[i]==3)scanf("%d%d",l+i,k+i);
 93         else scanf("%d%d%d",l+i,r+i,k+i);
 94         if(t[i]!=2)Q[n+i]=mp(k[i],i+n),ans[i]=1;
 95     }
 96     sort(Q+1,Q+n+m+1);
 97     for(int i=1;i<=n+m;++i){
 98         if(i==1||Q[i].first!=Q[i-1].first)back[++O]=Q[i].first;
 99         if(Q[i].second>n)k[Q[i].second-n]=O;
100         else a[Q[i].second]=b[Q[i].second]=O;
101     }
102     for(int i=1;i<=n;++i)I(2,i,0,a[i],i);
103     for(int i=1;i<=m;++i){
104         if(t[i]==3)I(3,l[i],0,b[l[i]],i),I(2,l[i],0,b[l[i]]=k[i],i);
105         else if(t[i]!=2){
106             if(t[i]==5)++k[i];
107             else if(t[i]==4)ans[i]=0;
108             I(1,r[i],0,k[i],i);
109             if(l[i]>1)I(0,l[i]-1,0,k[i],i);
110         }
111         else ans[i]=k[i];
112     }
113     cdq(1,P);
114     P=0;
115     for(int i=1;i<=n;++i)I(2,i,0,a[i],i),b[i]=a[i];
116     for(int i=1;i<=m;++i){
117         if(t[i]==3)I(3,l[i],0,b[l[i]],i),I(2,l[i],0,b[l[i]]=k[i],i);
118         else if(t[i]!=1)I(1,l[i],r[i],ans[i],i);
119     }
120     go(1,O+1,1,P);
121     for(int i=1;i<=m;++i)if(t[i]!=3)printf("%d
",ans[i]);
122     return 0;
123 }
    View Code

    主席树

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct T{
 8     int l,r,s;
 9 }t[4000005];
10 int n,m,cnt,r[100005];
11 void add(int l,int r,int x,int y,int k){
12     t[x]=t[y],++t[x].s;
13     if(l==r) return;
14     int m=(l+r)>>1;
15     if(k<=m)add(l,m,t[x].l=++cnt,t[y].l,k);
16     else add(m+1,r,t[x].r=++cnt,t[y].r,k);
17 }
18 int get(int l,int r,int x,int y,int k){
19     if(l==r) return l;
20     int m=(l+r)>>1;
21     if(t[t[x].l].s-t[t[y].l].s>=k) return get(l,m,t[x].l,t[y].l,k);
22     return get(m+1,r,t[x].r,t[y].r,k-t[t[x].l].s+t[t[y].l].s);
23 }
24 int main(){
25     scanf("%d%d",&n,&m);
26     for(int i=1;i<=n;++i){
27         int x;
28         scanf("%d",&x);
29         add(-orz,orz,r[i]=++cnt,r[i-1],x);
30     }
31     for(int i=1;i<=m;++i){
32         int x,y,k;
33         scanf("%d%d%d",&x,&y,&k);
34         printf("%d
",get(-orz,orz,r[y],r[x-1],k));
35     }
36     return 0;
37 }
    View Code

    KMP

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int a[1000005],b[1000005],n,m,f[1000005];
 8 string s1,s2;
 9 int main(){
10     ios::sync_with_stdio(0);
11     cin>>s1>>s2;
12     n=s1.size(),m=s2.size();
13     for(int i=1;i<=n;++i)a[i]=s1[i-1];
14     for(int i=1;i<=m;++i)b[i]=s2[i-1];
15     int k=0;
16     for(int i=2;i<=m;++i){
17         while(b[k+1]!=b[i]&&k)k=f[k];
18         if(b[k+1]==b[i])++k;
19         f[i]=k;
20     }
21     k=0;
22     for(int i=1;i<=n;++i){
23         while(b[k+1]!=a[i]&&k)k=f[k];
24         if(b[k+1]==a[i])++k;
25         if(k==m)printf("%d
",i-m+1);
26     }
27     for(int i=1;i<m;++i)printf("%d ",f[i]);
28     printf("%d
",f[m]);
29     return 0;
30 }
    View Code

    manacher

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int a[22000005],f[22000005],n,k,ans;
 8 int main()
 9 {
10     a[0]='+';
11     a[++n]='?';
12     char c=getchar();
13     while(c>='a'){
14                    a[++n]=c;
15                    a[++n]='?';
16                    c=getchar();
17     }
18     a[n+1]='-';
19     for(int i=1;i<=n;i++){
20             if(i<k+f[k])f[i]=min(f[k]+k-i,f[k+k-i]);
21             while(a[i-f[i]]==a[i+f[i]])f[i]++;
22             if(f[i]+i>f[k]+k)k=i;
23     }
24     for(int i=1;i<=n;i++)ans=max(ans,f[i]);
25     printf("%d
",ans-1);
26     return 0;
27 }
    View Code

    后缀数组

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int n,a[100005],cnt[200],r[240005],R[100005],p[100005],q[100005],C[100005],sa[100005],h[100005];
 8 int main(){
 9     char c=getchar();
10     while(c!='
')a[++n]=c,cnt[c]=1,c=getchar();
11     for(int i=40;i<=150;++i)cnt[i]+=cnt[i-1];
12     for(int i=1;i<=n;++i)r[i]=cnt[a[i]];
13     for(int i=2;i<=n;++i)++C[r[i]];
14     C[0]=1;
15     for(int i=1;i<=n;++i)C[i]+=C[i-1];
16     for(int i=n;i;--i)q[C[r[i+1]]--]=i;
17     memset(C,0,sizeof(C));
18     for(int _=1;_<n;_<<=1){
19         int __=_<<1;
20         for(int i=1;i<=n;++i)++C[r[i]];
21         for(int i=1;i<=n;++i)C[i]+=C[i-1];
22         for(int i=n;i;--i)p[C[r[q[i]]]--]=q[i];
23         memset(C,0,sizeof(C));
24         int k=0,K=__;
25         if(__<n){
26             for(int i=1;i<=__;++i)q[i]=n+1-i;
27         }
28         for(int i=1;i<=n;++i){
29             if(r[p[i]]!=r[p[i-1]]||r[p[i]+_]!=r[p[i-1]+_])++k;
30             R[p[i]]=k;
31             if(p[i]>__)q[++K]=p[i]-__;
32         }
33         for(int i=1;i<=n;++i)r[i]=R[i];
34     }
35     for(int i=1;i<=n;++i)sa[r[i]]=i;
36     for(int i=1;i<n;++i)printf("%d ",sa[i]);
37     printf("%d
",sa[n]);
38     int k=0;
39     for(int i=1;i<=n;++i){
40         if(r[i]==1){
41             k=0;
42             continue;
43         }
44         if(k)--k;
45         int j=sa[r[i]-1];
46         while(a[i+k]==a[j+k])++k;
47         h[r[i]]=k;
48     }
49     for(int i=2;i<n;++i)printf("%d ",h[i]);
50     if(n>1)printf("%d
",h[n]);
51     return 0;
52 }
    View Code

    凸包

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 #define X    first
 7 #define Y    second
 8 using namespace std;
 9 struct P{int x,y;};
10 P mns(P x,P y){x.x-=y.x,x.y-=y.y;return x;}
11 int det(P x,P y){return x.x*y.y-x.y*y.x;}
12 int norm(P x){return x.x*x.x+x.y*x.y;}
13 int n;
14 pair<int,int> p[50005];
15 P a[50005],b[50005],c[50005];
16 int lb,lc;
17 int ans;
18 int main(){
19     scanf("%d",&n);
20     for(int i=1;i<=n;++i)scanf("%d%d",&p[i].X,&p[i].Y);
21     sort(p+1,p+n+1);
22     for(int i=1;i<=n;++i)a[i].x=p[i].X,a[i].y=p[i].Y;
23     for(int i=1;i<=n;++i){
24         while(lb>1&&det(mns(b[lb],b[lb-1]),mns(a[i],b[lb]))>=0)--lb;
25         b[++lb]=a[i];
26     }
27     for(int i=1;i<=n;++i){
28         while(lc>1&&det(mns(c[lc],c[lc-1]),mns(a[i],c[lc]))<=0)--lc;
29         c[++lc]=a[i];
30     }
31     for(int i=1;i<=lb;++i){
32         for(int j=1;j<=lc;++j)ans=max(ans,norm(mns(b[i],c[j])));
33     }
34     printf("%d
",ans);
35     return 0;
36 }
    View Code

     一般图最大匹配(带花树)

    #include<bits/stdc++.h>
#define ll   long long
#define pb   push_back
#define mp   make_pair
#define orz  1000000007
using namespace std;
int n,m,mat[505],pre[505],ans,cnt,a[505],fa[505],c[505],q[505],ql,qr;
vector<int> v[505];
int get(int x){return x==fa[x]?x:fa[x]=get(fa[x]);}
int lca(int x,int y){
    ++cnt;
    x=get(x),y=get(y);
    while(1){
        if(x){
            if(a[x]==cnt) return x;
            a[x]=cnt,x=get(pre[mat[x]]);
        }
        swap(x,y);
    }
}
void bl(int x,int y,int z){
    while(get(x)!=z){
        pre[x]=y;
        if(c[mat[x]]==1)q[++qr]=mat[x],c[mat[x]]=0;
        if(fa[x]==x)fa[x]=z;
        if(fa[mat[x]]==mat[x])fa[mat[x]]=z;
        y=mat[x],x=pre[y];
    }
}
bool go(int x){
    for(int i=1;i<=n;++i)fa[i]=i;
    memset(c,-1,sizeof(c));
    c[x]=0,qr=1,ql=0,q[1]=x;
    while(ql<qr){
        int o=q[++ql];
        for(int j=0;j<v[o].size();++j){
            int p=v[o][j];
            if(c[p]==-1){
                pre[p]=o,c[p]=1;
                if(!mat[p]){
                    while(o){
                        int t=mat[o];
                        mat[o]=p,mat[p]=o;
                        p=t,o=pre[t];
                    }
                    return 1;
                }
                q[++qr]=mat[p],c[mat[p]]=0;
            }
            else if(!c[p]){
                int t=lca(o,p);
                bl(o,p,t),bl(p,o,t);
            }
        }
    }
    return 0;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i){
        int x,y;
        scanf("%d%d",&x,&y);
        v[x].pb(y),v[y].pb(x);
    }
    for(int i=1;i<=n;++i)if(!mat[i]&&go(i))++ans;
    printf("%d
",ans);
    for(int i=1;i<n;++i)printf("%d ",mat[i]);
    printf("%d
",mat[n]);
    return 0;
}
    View Code

    一般图最大匹配(乱搞)

    #include<bits/stdc++.h>
#define ll   long long
#define pb   push_back
#define mp   make_pair
#define orz  1000000007
using namespace std;
int n,m,a[505],u[505],ans[505],res,p[505];
vector<int> v[505];
bool dfs(int x){
    u[x]=1;
    random_shuffle(v[x].begin(),v[x].end());
    for(int i=0;i<v[x].size();++i){
        int y=v[x][i];
        if(!a[y]){
            a[x]=y,a[y]=x;
            return 1;
        }
    }
    for(int i=0;i<v[x].size();++i){
        int y=v[x][i],z=a[y];
        if(u[z]) continue;
        a[x]=y,a[y]=x,a[z]=0;
        if(dfs(z)) return 1;
        a[x]=0,a[y]=z,a[z]=y;
    }
    return 0;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i){
        int x,y;
        scanf("%d%d",&x,&y);
        v[x].pb(y),v[y].pb(x);
    }
    for(int i=1;i<=n;++i)p[i]=i;
    for(int _=10;_;--_){
        random_shuffle(p+1,p+n+1);
        for(int i=1;i<=n;++i){
            if(a[p[i]]) continue;
            memset(u,0,sizeof(u));
            dfs(p[i]);
        }
        int cnt=0;
        for(int i=1;i<=n;++i)if(a[i])++cnt;
        if(cnt/2>res){
            res=cnt/2;
            for(int i=1;i<=n;++i)ans[i]=a[i];
        }
    }
    printf("%d
",res);
    for(int i=1;i<n;++i)printf("%d ",ans[i]);
    printf("%d
",ans[n]);        
    return 0;
}
    View Code

    二分图最大权匹配

    #include<bits/stdc++.h>
#define ll   long long
#define pb   push_back
#define mp   make_pair
#define orz  1000000007
using namespace std;
int nl,nr,n,m,f[405][405],lx[405],ly[405],vx[405],vy[405],sl[405];
int q[405],ql,qr,pre[405],ml[405],mr[405],out[405];
void go(int x){
    while(1){
        ql=0,qr=1,vx[x]=1,q[1]=x;
        while(ql<qr){
            int o=q[++ql];
            for(int i=1;i<=n;++i){
                if(vy[i]) continue;
                int t=lx[o]+ly[i]-f[o][i];
                if(t>sl[i]) continue;
                pre[i]=o;
                if(!t){
                    if(!ml[i]){
                        while(i){
                            ml[i]=pre[i];
                            swap(i,mr[pre[i]]);
                        }
                        return;
                    }
                    q[++qr]=ml[i];
                    vy[i]=vx[ml[i]]=1;
                }
                else sl[i]=t;
            }
        }
        int o,O=orz;
        for(int i=1;i<=n;++i)if(!vy[i]&&O>sl[i])o=i,O=sl[i];
        for(int i=1;i<=n;++i){
            if(vx[i])lx[i]-=O;
            if(vy[i])ly[i]+=O;
            else sl[i]-=O;
        }
        if(!ml[o]){
            while(o){
                ml[o]=pre[o];
                swap(o,mr[pre[o]]);
            }
            return;
        }
        vy[o]=1,x=ml[o];
    }
}        
int main(){
    scanf("%d%d%d",&nl,&nr,&m);
    n=max(nl,nr);
    for(int i=1;i<=m;++i){
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        f[x][y]=z;
        lx[x]=max(lx[x],z);
    }
    for(int i=1;i<=n;++i){
        memset(vx,0,sizeof(vx));
        memset(vy,0,sizeof(vy));
        for(int j=1;j<=n;++j)sl[j]=orz;
        go(i);
    }
    ll ans=0;
    for(int i=1;i<=n;++i)ans+=lx[i]+ly[i];
    printf("%lld
",ans);
    for(int i=1;i<=nl;++i)out[i]=f[i][mr[i]]?mr[i]:0;
    for(int i=1;i<nl;++i)printf("%d ",out[i]);
    printf("%d
",out[nl]);
    return 0;
}
    View Code

     多项式求逆

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define M    998244353ll
 6 using namespace std;
 7 struct T{
 8     int n;
 9     ll a[524444],b[524444],c[524444],w[524444],w2[524444],r;
10     ll P(ll x,ll k){
11         ll ans=1;
12         while(k){
13             if(k&1ll)ans=ans*x%M;
14             k>>=1,x=x*x%M;
15         }
16         return ans;
17     }
18     void ntt(ll *a,ll *w){
19         for(int i=0,j=0;i<n;++i){
20             if(i<j)swap(a[i],a[j]);
21             for(int k=n>>1;(j^=k)<k;k>>=1);
22         }
23         for(int i=2;i<=n;i<<=1){
24             for(int j=0;j<n;j+=i){
25                 for(int k=0;k<(i>>1);++k){
26                     int l=j+k,r=l+(i>>1);
27                     ll o=a[r]*w[n/i*k]%M;
28                     a[r]=(a[l]+M-o)%M,a[l]=(a[l]+o)%M;
29                 }
30             }
31         }
32     }          
33     void init(){
34         r=P(n,M-2),w[0]=w2[0]=1,w[1]=w2[n-1]=P(3,(M-1)/n);
35         for(int i=2;i<n;++i)w[i]=w2[n-i]=w[i-1]*w[1]%M;
36     }
37     void mul(){
38         init();
39         ntt(a,w),ntt(b,w);
40         for(int i=0;i<n;++i)c[i]=a[i]*b[i]%M*r%M;
41         ntt(c,w2);
42     }
43 }t;
44 int n;
45 ll a[266666],b[266666];
46 int main(){
47     scanf("%d",&n);
48     for(int i=0;i<n;++i)scanf("%d",a+i);
49     b[0]=t.P(a[0],M-2ll);
50     for(int d=2;d<=262144;d<<=1){
51         t.n=d;
52         for(int i=0;i<(d>>1);++i)t.a[i]=a[i],t.b[i]=b[i];
53         for(int i=d>>1;i<d;++i)t.a[i]=t.b[i]=0;
54         t.mul();
55         t.n=d<<1;
56         for(int i=0;i<d;++i)t.a[i]=t.c[i],t.b[i]=b[i];
57         for(int i=d;i<(d<<1);++i)t.a[i]=t.b[i]=0;
58         t.mul();
59         for(int i=0;i<d;++i)b[i]=(b[i]*2ll-t.c[i]+M)%M;
60     }
61     for(int i=0;i<n-1;++i)printf("%lld ",b[i]);
62     printf("%lld
",b[n-1]);
63     return 0;
64 }
    View Code

     杜教筛

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int T,n,f[5000005],p[800005],P;
 8 ll pha[5000005],phb[444],mua[5000005],mub[444];
 9 bool PHB[444],MUB[444];
10 ll getph(int x){
11     if(x<=5000000) return pha[x];
12     int y=n/x;
13     if(PHB[y]&&n/y==x) return phb[y];
14     ll ret=x*(x+1ll)/2ll;
15     for(int l=2,r;l<=x;l=r+1){
16         r=x/(x/l);
17         ret-=(r-l+1ll)*getph(x/l);
18     }
19     if(n/y==x)PHB[y]=1,phb[y]=ret;
20     return ret;
21 }
22 ll getmu(int x){
23     if(x<=5000000) return mua[x];
24     int y=n/x;
25     if(MUB[y]&&n/y==x) return mub[y];
26     ll ret=1;
27     for(int l=2,r;l<=x;l=r+1){
28         r=x/(x/l);
29         ret-=(r-l+1ll)*getmu(x/l);
30     }
31     if(n/y==x)MUB[y]=1,mub[y]=ret;
32     return ret;
33 }
34 int main(){
35     pha[1]=mua[1]=1;
36     for(int i=2;i<=5000000;++i){
37         if(!f[i])p[++P]=i;
38         for(int j=1;j<=P&&p[j]<=5000000/i;++j){
39             f[p[j]*i]=p[j];
40             if(i%p[j]==0) break;
41         }
42     }
43     for(int i=2;i<=5000000;++i){
44         if(!f[i])pha[i]=i-1,mua[i]=-1;
45         else{
46             int o=i/f[i];
47             if(o%f[i]==0)pha[i]=pha[o]*f[i],mua[i]=0;
48             else pha[i]=pha[o]*(f[i]-1),mua[i]=-mua[o];
49         }
50     }
51     for(int i=2;i<=5000000;++i)pha[i]+=pha[i-1],mua[i]+=mua[i-1];
52     scanf("%d",&T);
53     while(T--){
54         memset(phb,0,sizeof(phb));
55         memset(mub,0,sizeof(mub));
56         memset(PHB,0,sizeof(PHB));
57         memset(MUB,0,sizeof(MUB));
58         scanf("%d",&n);
59         printf("%lld %lld
",getph(n),getmu(n));
60     } 
61     return 0;
62 }
    View Code

     任意模数NTT

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define C    pair<long double,long double>
 6 #define X    first
 7 #define Y    second
 8 using namespace std;
 9 struct MTT{
10     int n,m,A[262222],B[262222],ans[262222];
11     ll P;
12     C w[262222],w2[262222],a[262222],b[262222],a2[262222],b2[262222];
13     C mul(C &x,C &y){
14         return mp(x.X*y.X-x.Y*y.Y,x.X*y.Y+x.Y*y.X);
15     }
16     void gen(){
17         long double pi=acos(-1.0);
18         w2[0]=w[0]=mp(1.0,0.0);
19         for(int i=1;i<n;++i)w2[n-i]=w[i]=mp(cos(2.0*i*pi/n),sin(2.0*i*pi/n));
20     }
21     void fft(C *a,C *w){
22         for(int i=0,j=0;i<n;++i){
23             if(i<j)swap(a[i],a[j]);
24             for(int k=n>>1;(j^=k)<k;k>>=1);
25         }
26         for(int i=2;i<=n;i<<=1){
27             for(int j=0;j<n;j+=i){
28                 for(int k=0;k<(i>>1);++k){
29                     int l=j+k,r=l+(i>>1);
30                     C o=mul(a[r],w[n/i*k]);
31                     a[r].X=a[l].X-o.X,a[r].Y=a[l].Y-o.Y,a[l].X+=o.X,a[l].Y+=o.Y;
32                 }
33             }
34         }
35     }
36     void mtt(){
37         gen();
38         for(int i=0;i<n;++i)a[i]=mp(A[i]&32767,A[i]>>15),b[i]=mp(B[i]&32767,B[i]>>15);
39         fft(a,w),fft(b,w);
40         for(int i=0;i<n;++i){
41             int j=(i?n-i:0);
42             C aa,bb,cc,dd,ac,ad,bc,bd;
43             aa=mp((a[i].X+a[j].X)*0.5,(a[i].Y-a[j].Y)*0.5);
44             bb=mp((a[i].Y+a[j].Y)*0.5,(a[j].X-a[i].X)*0.5);
45             cc=mp((b[i].X+b[j].X)*0.5,(b[i].Y-b[j].Y)*0.5);
46             dd=mp((b[i].Y+b[j].Y)*0.5,(b[j].X-b[i].X)*0.5);
47             ac=mul(aa,cc),ad=mul(aa,dd),bc=mul(bb,cc),bd=mul(bb,dd);
48             a2[j]=mp(ac.X-ad.Y,ac.Y+ad.X),b2[j]=mp(bc.X-bd.Y,bc.Y+bd.X);
49         }
50         fft(a2,w),fft(b2,w);
51         for(int i=0;i<n;++i){
52             ll aa,bb,cc,dd;
53             aa=a2[i].X/n+0.5,bb=a2[i].Y/n+0.5,cc=b2[i].X/n+0.5,dd=b2[i].Y/n+0.5;
54             aa%=P,bb%=P,cc%=P,dd%=P;
55             ll o=aa+(bb+cc)*32768ll%P+dd*32768ll%P*32768ll%P;
56             o=(o%P+P)%P;
57             ans[i]=o;
58         }
59     }
60     void solve(){
61         for(int i=0;i<=n;++i)A[i]=(A[i]%P+P)%P;
62         for(int i=0;i<=m;++i)B[i]=(B[i]%P+P)%P;
63         m+=n,n=1;
64         while(n<=m)n<<=1;
65         mtt();
66     }
67 }_;
68 int main(){
69     scanf("%d%d%lld",&_.n,&_.m,&_.P);
70     for(int i=0;i<=_.n;++i)scanf("%d",_.A+i);
71     for(int i=0;i<=_.m;++i)scanf("%d",_.B+i);
72     _.solve();
73     for(int i=0;i<_.m;++i)printf("%d ",_.ans[i]);
74     printf("%d
",_.ans[_.m]);
75     return 0;
76 }
    View Code

     最小树形图

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int n,m,r,x[10005],y[10005],z[10005],d[105],u[105],id[105],pre[105];
 8 int solve(){
 9     int N=n,ans=0;
10     while(N){
11         for(int i=1;i<=n;++i)d[i]=orz,u[i]=0,id[i]=0,pre[i]=0;
12         for(int i=1;i<=m;++i){
13             if(x[i]!=y[i]&&d[y[i]]>z[i])d[y[i]]=z[i],pre[y[i]]=x[i];
14         }
15         pre[r]=r,d[r]=0;
16         for(int i=1;i<=n;++i){
17             if(!pre[i]) return -1;
18             ans+=d[i];
19         }
20         N=0;
21         for(int i=1;i<=n;++i){
22             if(u[i]) continue;
23             int I=i;
24             while(!u[I])u[I]=i,I=pre[I];
25             if(u[I]!=i||I==r) continue;
26             ++N;
27             while(!id[I])id[I]=N,I=pre[I];
28         }
29         if(!N) return ans;
30         for(int i=1;i<=n;++i)if(!id[i])id[i]=++N;
31         for(int i=1;i<=m;++i)z[i]-=d[y[i]],x[i]=id[x[i]],y[i]=id[y[i]];
32         n=N,r=id[r];
33     }
34 }
35 int main(){
36     scanf("%d%d%d",&n,&m,&r);
37     for(int i=1;i<=m;++i)scanf("%d%d%d",x+i,y+i,z+i);
38     printf("%d
",solve());
39     return 0;
40 }
    View Code

     BM-CH

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct BM{
 8     int n,cnt,f[2005];
 9     ll a[2005],d[2005];
10     vector<ll> v[2005];
11     ll fp(ll x,ll k=1000000005){
12         ll ans=1;
13         while(k){
14             if(k&1)ans=ans*x%orz;
15             k>>=1,x=x*x%orz;
16         }
17         return ans;
18     }
19     void solve(){
20         while(cnt)v[cnt--].clear();
21         v[0].clear();
22         for(int i=1;i<=n;++i){
23             ll t=a[i];
24             for(int j=0;j<v[cnt].size();++j)(t-=v[cnt][j]*a[i-j-1])%=orz;
25             t=(t%orz+orz)%orz;
26             d[i]=t;
27             if(!t) continue;
28             f[cnt]=i;
29             if(!cnt){
30                 v[++cnt].resize(i);
31                 continue;
32             }
33             ll mul=d[i]*fp(d[f[cnt-1]])%orz;
34             ++cnt;
35             v[cnt].resize(i-f[cnt-2]-1);
36             v[cnt].pb(mul);
37             for(int j=0;j<v[cnt-2].size();++j){
38                 ll o=-mul*v[cnt-2][j];
39                 o=(o%orz+orz)%orz;
40                 v[cnt].pb(o);
41             }
42             if(v[cnt].size()<v[cnt-1].size())v[cnt].resize(v[cnt-1].size());
43             for(int j=0;j<v[cnt-1].size();++j){
44                 v[cnt][j]+=v[cnt-1][j];
45                 if(v[cnt][j]>=orz)v[cnt][j]-=orz;
46             }
47         }
48     }
49 }b;
50 struct CH{
51     int n,k;
52     ll f[2005],a[2005],g[2005],ans[2005],x[2005],res[4005];
53     void add(ll &x,ll y){
54         x+=y;
55         if(x>=orz)x-=orz;
56     }
57     void mul(ll *A,ll *B,ll *C){
58         for(int i=0;i<=k*2-2;++i)res[i]=0;
59         for(int i=0;i<k;++i){
60             for(int j=0;j<k;++j)(res[i+j]+=A[i]*B[j])%=orz;
61         }
62         for(int i=k*2-2;i>=k;--i){
63             ll o=res[i];
64             if(!o) continue;
65             for(int j=0;j<k;++j)add(res[i-k+j],orz-o*g[j]%orz);
66         }
67         for(int i=0;i<k;++i)C[i]=res[i];
68     }
69     ll solve(){
70         if(k==1){
71             return f[0]*b.fp(a[0],n-k)%orz;
72         }
73         g[k]=1;
74         for(int i=1;i<=k;++i)g[k-i]=(orz-a[i-1])%orz;
75         for(int i=0;i<k;++i)ans[i]=x[i]=0;
76         ans[0]=1,x[1]=1;
77         int T=n;
78         while(T){
79             if(T&1)mul(ans,x,ans);
80             T>>=1,mul(x,x,x);
81         }
82         ll ret=0;
83         for(int i=0;i<k;++i)(ret+=ans[i]*f[i])%=orz;
84         return (ret%orz+orz)%orz;
85     }
86 }c;
87 int main(){
88     scanf("%d%d",&b.n,&c.n);
89     --c.n;
90     for(int i=1;i<=b.n;++i)scanf("%lld",b.a+i);
91     b.solve();
92     c.k=b.v[b.cnt].size();
93     for(int i=0;i<c.k;++i)c.f[i]=b.a[i+1],c.a[i]=b.v[b.cnt][i];
94     printf("%lld
",c.solve());
95     return 0;
96 }
    View Code

     无源汇上下界可行流

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct E{int x,y,z;};
 8 struct F{
 9     int n,s,t,l[205],i[205],q[205],ql,qr;
10     vector<E> v[205];
11     void add(int x,int y,int z){
12         E X,Y;
13         X.x=y,X.y=z,X.z=v[y].size();
14         Y.x=x,Y.y=0,Y.z=v[x].size();
15         v[x].pb(X),v[y].pb(Y);
16     }
17     void bfs(){
18         memset(l,-1,sizeof(l));
19         l[q[qr=1]=s]=0;
20         for(ql=1;ql<=qr;++ql){
21             int x=q[ql];
22             for(int i=0;i<v[x].size();++i)if(l[v[x][i].x]==-1&&v[x][i].y)l[q[++qr]=v[x][i].x]=l[x]+1;
23         }
24     }
25     int dfs(int x,int y){
26         if(x==t) return y;
27         int ret=0;
28         for(;i[x]<v[x].size();++i[x]){
29             E &e=v[x][i[x]];
30             if(!e.y||l[e.x]!=l[x]+1) continue;
31             int d=dfs(e.x,min(e.y,y-ret));
32             if(!d) continue;
33             ret+=d,e.y-=d,v[e.x][e.z].y+=d;
34             if(ret==y) return ret;
35         }
36         l[x]=-1;
37         return ret;
38     }
39     int dinic(){
40         int ans=0;
41         while(1){
42             bfs();
43             if(l[t]==-1) return ans;
44             memset(i,0,sizeof(i));
45             ans+=dfs(s,orz);
46         }
47     }
48 }_;
49 int n,m,s[11000],t[11000],l[11000],r[11000],S[205],T[205],res,rec[11000];
50 int main(){
51     scanf("%d%d",&n,&m);
52     _.n=_.t=n+2,_.s=n+1;
53     for(int i=1;i<=m;++i){
54         scanf("%d%d%d%d",s+i,t+i,l+i,r+i);
55         if(l[i]>r[i]){
56             puts("NO");
57             return 0;
58         }
59         rec[i]=_.v[t[i]].size();
60         _.add(s[i],t[i],r[i]-l[i]);
61         S[t[i]]+=l[i],T[s[i]]+=l[i],res+=l[i];
62     }
63     for(int i=1;i<=n;++i)_.add(_.s,i,S[i]),_.add(i,_.t,T[i]);
64     res-=_.dinic();
65     if(res)puts("NO");
66     else{
67         puts("YES");
68         for(int i=1;i<=m;++i)printf("%d
",l[i]+_.v[t[i]][rec[i]].y);
69     }
70     return 0;
71 }
    View Code

    有源汇上下界最大流

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct E{int x,y,z;};
 8 struct F{
 9     int n,s,t,l[205],i[205],q[205],ql,qr;
10     vector<E> v[205];
11     void add(int x,int y,int z){
12         E X,Y;
13         X.x=y,X.y=z,X.z=v[y].size();
14         Y.x=x,Y.y=0,Y.z=v[x].size();
15         v[x].pb(X),v[y].pb(Y);
16     }
17     void bfs(){
18         memset(l,-1,sizeof(l));
19         l[q[qr=1]=s]=0;
20         for(ql=1;ql<=qr;++ql){
21             int x=q[ql];
22             for(int i=0;i<v[x].size();++i)if(l[v[x][i].x]==-1&&v[x][i].y)l[q[++qr]=v[x][i].x]=l[x]+1;
23         }
24     }
25     int dfs(int x,int y){
26         if(x==t) return y;
27         int ret=0;
28         for(;i[x]<v[x].size();++i[x]){
29             E &e=v[x][i[x]];
30             if(!e.y||l[e.x]!=l[x]+1) continue;
31             int d=dfs(e.x,min(e.y,y-ret));
32             if(!d) continue;
33             ret+=d,e.y-=d,v[e.x][e.z].y+=d;
34             if(ret==y) return ret;
35         }
36         l[x]=-1;
37         return ret;
38     }
39     int dinic(){
40         int ans=0;
41         while(1){
42             bfs();
43             if(l[t]==-1) return ans;
44             memset(i,0,sizeof(i));
45             ans+=dfs(s,orz);
46         }
47     }
48 }_,f;
49 int n,m,ss,tt,s[11000],t[11000],l[11000],r[11000],S[205],T[205],res,rec[11000];
50 int main(){
51     scanf("%d%d%d%d",&n,&m,&ss,&tt);
52     _.n=_.t=n+2,_.s=n+1;
53     for(int i=1;i<=m;++i){
54         scanf("%d%d%d%d",s+i,t+i,l+i,r+i);
55         if(l[i]>r[i]){
56             puts("please go home to sleep");
57             return 0;
58         }
59         rec[i]=_.v[t[i]].size();
60         _.add(s[i],t[i],r[i]-l[i]);
61         S[t[i]]+=l[i],T[s[i]]+=l[i],res+=l[i];
62     }
63     for(int i=1;i<=n;++i)_.add(_.s,i,S[i]),_.add(i,_.t,T[i]);
64     _.add(tt,ss,orz);
65     res-=_.dinic();
66     if(res){
67         puts("please go home to sleep");
68         return 0;
69     }
70     for(int i=0;i<_.v[ss].size();++i)if(_.v[ss][i].x==tt)res=_.v[ss][i].y;
71     f.n=n,f.s=ss,f.t=tt;
72     for(int i=1;i<=m;++i){
73         int o=_.v[t[i]][rec[i]].y;
74         E X,Y;
75         X.x=t[i],X.y=r[i]-l[i]-o,X.z=f.v[t[i]].size();
76         Y.x=s[i],Y.y=o,Y.z=f.v[s[i]].size();
77         f.v[s[i]].pb(X),f.v[t[i]].pb(Y);
78     }
79     printf("%d
",res+f.dinic());
80     return 0;
81 }
    View Code

    有源汇上下界最小流

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  2147483647
 6 using namespace std;
 7 struct E{int x,y,z;};
 8 struct F{
 9     int n,s,t,l[50010],i[50010],q[50010],ql,qr;
10     vector<E> v[50010];
11     void add(int x,int y,int z){
12         E X,Y;
13         X.x=y,X.y=z,X.z=v[y].size();
14         Y.x=x,Y.y=0,Y.z=v[x].size();
15         v[x].pb(X),v[y].pb(Y);
16     }
17     void bfs(){
18         memset(l,-1,sizeof(l));
19         l[q[qr=1]=s]=0;
20         for(ql=1;ql<=qr;++ql){
21             int x=q[ql];
22             for(int i=0;i<v[x].size();++i)if(l[v[x][i].x]==-1&&v[x][i].y)l[q[++qr]=v[x][i].x]=l[x]+1;
23         }
24     }
25     int dfs(int x,int y){
26         if(x==t) return y;
27         int ret=0;
28         for(;i[x]<v[x].size();++i[x]){
29             E &e=v[x][i[x]];
30             if(!e.y||l[e.x]!=l[x]+1) continue;
31             int d=dfs(e.x,min(e.y,y-ret));
32             if(!d) continue;
33             ret+=d,e.y-=d,v[e.x][e.z].y+=d;
34             if(ret==y) return ret;
35         }
36         l[x]=-1;
37         return ret;
38     }
39     int dinic(){
40         int ans=0;
41         while(1){
42             bfs();
43             if(l[t]==-1) return ans;
44             memset(i,0,sizeof(i));
45             ans+=dfs(s,orz);
46         }
47     }
48 }_,f;
49 int n,m,ss,tt,s[125010],t[125010],l[125010],r[125010],S[50010],T[50010],res,rec[125010];
50 int main(){
51     scanf("%d%d%d%d",&n,&m,&ss,&tt);
52     _.n=_.t=n+2,_.s=n+1;
53     for(int i=1;i<=m;++i){
54         scanf("%d%d%d%d",s+i,t+i,l+i,r+i);
55         if(l[i]>r[i]){
56             puts("please go home to sleep");
57             return 0;
58         }
59         rec[i]=_.v[t[i]].size();
60         _.add(s[i],t[i],r[i]-l[i]);
61         S[t[i]]+=l[i],T[s[i]]+=l[i],res+=l[i];
62     }
63     for(int i=1;i<=n;++i)_.add(_.s,i,S[i]),_.add(i,_.t,T[i]);
64     _.add(tt,ss,orz);
65     res-=_.dinic();
66     if(res){
67         puts("please go home to sleep");
68         return 0;
69     }
70     for(int i=0;i<_.v[ss].size();++i)if(_.v[ss][i].x==tt)res=_.v[ss][i].y;
71     f.n=n,f.s=tt,f.t=ss;
72     for(int i=1;i<=m;++i){
73         int o=_.v[t[i]][rec[i]].y;
74         E X,Y;
75         X.x=t[i],X.y=r[i]-l[i]-o,X.z=f.v[t[i]].size();
76         Y.x=s[i],Y.y=o,Y.z=f.v[s[i]].size();
77         f.v[s[i]].pb(X),f.v[t[i]].pb(Y);
78     }
79     printf("%d
",res-f.dinic());
80     return 0;
81 }
    View Code

     KDT

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 struct T{int l,r,x[2],d[2],k[2],s,v;}t[200005];
 8 int N,n,ans,D,r=1,X[2],Y[2];
 9 bool cmp(T x,T y){return x.k[D]==y.k[D]?x.k[D^1]<y.k[D^1]:x.k[D]<y.k[D];}
10 void upd(int k){
11     t[k].x[0]=t[k].d[0]=t[k].k[0];
12     t[k].x[1]=t[k].d[1]=t[k].k[1];
13     t[k].s=t[k].v;
14     if(t[k].l){
15         t[k].s+=t[t[k].l].s;
16         if(t[t[k].l].x[0]<t[k].x[0])t[k].x[0]=t[t[k].l].x[0];
17         if(t[t[k].l].x[1]<t[k].x[1])t[k].x[1]=t[t[k].l].x[1];
18         if(t[t[k].l].d[0]>t[k].d[0])t[k].d[0]=t[t[k].l].d[0];
19         if(t[t[k].l].d[1]>t[k].d[1])t[k].d[1]=t[t[k].l].d[1];
20     }
21     if(t[k].r){
22         t[k].s+=t[t[k].r].s;
23         if(t[t[k].r].x[0]<t[k].x[0])t[k].x[0]=t[t[k].r].x[0];
24         if(t[t[k].r].x[1]<t[k].x[1])t[k].x[1]=t[t[k].r].x[1];
25         if(t[t[k].r].d[0]>t[k].d[0])t[k].d[0]=t[t[k].r].d[0];
26         if(t[t[k].r].d[1]>t[k].d[1])t[k].d[1]=t[t[k].r].d[1];
27     }
28 }
29 int build(int l,int r,int d){
30     int m=(l+r)>>1;
31     D=d;
32     nth_element(t+l+1,t+m+1,t+r+1,cmp);
33     if(l<m)t[m].l=build(l,m-1,d^1);
34     if(m<r)t[m].r=build(m+1,r,d^1);
35     upd(m);
36     return m;
37 }
38 void ins(int k){
39     int o=r,d=0;
40     while(1){
41         t[o].s+=t[k].v;
42         if(t[k].x[0]<t[o].x[0])t[o].x[0]=t[k].x[0];
43         if(t[k].x[1]<t[o].x[1])t[o].x[1]=t[k].x[1];
44         if(t[k].d[0]>t[o].d[0])t[o].d[0]=t[k].d[0];
45         if(t[k].d[1]>t[o].d[1])t[o].d[1]=t[k].d[1];
46         if(mp(t[k].k[d],t[k].k[d^1])<mp(t[o].k[d],t[o].k[d^1])){
47             if(!t[o].l){t[o].l=k;return;}
48             else o=t[o].l;
49         }
50         else{
51             if(!t[o].r){t[o].r=k;return;}
52             else o=t[o].r;
53         }
54         d^=1;
55     }
56 }
57 void ask(int k){
58     if(t[k].x[0]>Y[0]||t[k].d[0]<X[0]||t[k].x[1]>Y[1]||t[k].d[1]<X[1]) return;
59     if(t[k].x[0]>=X[0]&&t[k].d[0]<=Y[0]&&t[k].x[1]>=X[1]&&t[k].d[1]<=Y[1]){
60         ans+=t[k].s;
61         return;
62     }
63     if(t[k].k[0]>=X[0]&&t[k].k[0]<=Y[0]&&t[k].k[1]>=X[1]&&t[k].k[1]<=Y[1])ans+=t[k].v;
64     if(t[k].l)ask(t[k].l);
65     if(t[k].r)ask(t[k].r);
66 }
67 int main(){
68     scanf("%d",&N);
69     while(1){
70         int T;
71         scanf("%d",&T);
72         if(T==1){
73             ++n;
74             scanf("%d%d%d",&t[n].k[0],&t[n].k[1],&t[n].v);
75             t[n].k[0]^=ans,t[n].k[1]^=ans,t[n].v^=ans;
76             upd(n);
77             if(!(n&16383)){
78                 for(int i=1;i<n;++i)t[i].l=t[i].r=0,upd(i);
79                 r=build(1,n,0);
80             }
81             else if(n>1)ins(n);
82         }
83         else if(T==2){
84             scanf("%d%d%d%d",X,X+1,Y,Y+1);
85             X[0]^=ans,X[1]^=ans,Y[0]^=ans,Y[1]^=ans;
86             ans=0;
87             ask(r);
88             printf("%d
",ans);
89         }
90         else break;
91     }
92     return 0;
93 }
    View Code

     判素数

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 ll x;
 8 ll mul(ll x,ll y,ll n){
 9     if(x<y)swap(x,y);
10     ll ans=0;
11     while(y){
12         if(y&1){
13             ans+=x;
14             if(ans>=n)ans-=n;
15         }
16         y>>=1;
17         x+=x;
18         if(x>=n)x-=n;
19     }
20     return ans;
21 }
22 ll fp(ll x,ll k,ll n){
23     ll ans=1;
24     while(k){
25         if(k&1)ans=mul(ans,x,n);
26         k>>=1,x=mul(x,x,n);
27     }
28     return ans;
29 }
30 bool ch(ll x,ll k){
31     ll e=(x-1)/2;
32     while(1){
33         ll o=fp(k,e,x);
34         if(o==x-1) return 1;
35         if(o==1){
36             if(e%2==0)e/=2;
37             else return 1;
38         }
39         else return 0;
40     }
41 }
42 bool ok(ll x){
43     if(!ch(x,2)) return 0;
44     if(!ch(x,3)) return 0;
45     if(!ch(x,5)) return 0;
46     if(!ch(x,7)) return 0;
47     if(!ch(x,11)) return 0;
48     if(!ch(x,13)) return 0;
49     if(!ch(x,17)) return 0;
50     if(!ch(x,19)) return 0;
51     if(!ch(x,101)) return 0;
52     return 1;
53 }
54 int f[1000005],p[30],k;
55 int main(){
56     f[1]=1;
57     for(int i=2;i<=1000000;++i){
58         if(f[i]) continue;
59         for(int j=i*2;j<=1000000;j+=i)f[j]=1;
60     }
61     for(int i=2;i<=100;++i)if(!f[i])p[++k]=i;
62     while(cin>>x){
63         if(x<=1000000){
64             if(f[x])puts("N");
65             else puts("Y");
66         }
67         else{
68             bool _=1;
69             for(int i=1;i<=k;++i)if(x%p[i]==0)_=0,i=k;
70             if(!_)puts("N");
71             else if(ok(x))puts("Y");
72             else puts("N");
73         }
74     }
75     return 0;
76 }
    View Code

     Lyndon 分解

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 string s;
 8 int n,a[1100000],l[1100000],m;
 9 int main(){
10     ios::sync_with_stdio(0);
11     cin>>s;
12     n=s.size();
13     for(int i=1;i<=n;++i)a[i]=s[i-1];
14     for(int i=1;i<=n;){
15         int j=i,k=i+1;
16         while(k<=n&&a[j]<=a[k]){
17             if(a[j]<a[k])j=i;
18             else ++j;
19             ++k;
20         }
21         while(i<=j)i+=k-j,l[++m]=i-1;
22     }
23     for(int i=1;i<m;++i)printf("%d ",l[i]);
24     printf("%d
",l[m]);
25     return 0;
26 }
    View Code

    fwt

      1 #include<bits/stdc++.h>
  2 #define ll   long long
  3 #define pb   push_back
  4 #define mp   make_pair
  5 #define orz  998244353
  6 using namespace std;
  7 int n,m,a[131111],b[131111];
  8 struct S{
  9     int a[131111],b[131111],c[131111];
 10     void mul(){
 11         for(int i=0;i<n;++i)c[i]=a[i]*1ll*b[i]%orz;
 12     }
 13     void out(){
 14         for(int i=0;i<n-1;++i)printf("%d ",c[i]);
 15         printf("%d
",c[n-1]);
 16     }
 17 }_o,_a,_x;
 18 void fwt_or(int *a){
 19     for(int i=2;i<=n;i<<=1){
 20         for(int j=0;j<n;j+=i){
 21             for(int k=0;k<(i>>1);++k){
 22                 int l=j+k,r=l+(i>>1);
 23                 a[r]+=a[l];
 24                 if(a[r]>=orz)a[r]-=orz;
 25             }
 26         }
 27     }
 28 }
 29 void ifwt_or(int *a){
 30     for(int i=2;i<=n;i<<=1){
 31         for(int j=0;j<n;j+=i){
 32             for(int k=0;k<(i>>1);++k){
 33                 int l=j+k,r=l+(i>>1);
 34                 a[r]+=orz-a[l];
 35                 if(a[r]>=orz)a[r]-=orz;
 36             }
 37         }
 38     }
 39 }
 40 void fwt_and(int *a){
 41     for(int i=2;i<=n;i<<=1){
 42         for(int j=0;j<n;j+=i){
 43             for(int k=0;k<(i>>1);++k){
 44                 int l=j+k,r=l+(i>>1);
 45                 a[l]+=a[r];
 46                 if(a[l]>=orz)a[l]-=orz;
 47             }
 48         }
 49     }
 50 }
 51 void ifwt_and(int *a){
 52     for(int i=2;i<=n;i<<=1){
 53         for(int j=0;j<n;j+=i){
 54             for(int k=0;k<(i>>1);++k){
 55                 int l=j+k,r=l+(i>>1);
 56                 a[l]+=orz-a[r];
 57                 if(a[l]>=orz)a[l]-=orz;
 58             }
 59         }
 60     }
 61 }
 62 void fwt_xor(int *a){
 63     for(int i=2;i<=n;i<<=1){
 64         for(int j=0;j<n;j+=i){
 65             for(int k=0;k<(i>>1);++k){
 66                 int l=j+k,r=l+(i>>1),x=a[l]+a[r],y=a[l]+orz-a[r];
 67                 a[l]=x,a[r]=y;
 68                 if(a[l]>=orz)a[l]-=orz;
 69                 if(a[r]>=orz)a[r]-=orz;
 70             }
 71         }
 72     }
 73 }
 74 void ifwt_xor(int *a){
 75     for(int i=2;i<=n;i<<=1){
 76         for(int j=0;j<n;j+=i){
 77             for(int k=0;k<(i>>1);++k){
 78                 int l=j+k,r=l+(i>>1),x=a[l]+a[r],y=a[l]+orz-a[r];
 79                 a[l]=x,a[r]=y;
 80                 if(a[l]>=orz)a[l]-=orz;
 81                 if(a[r]>=orz)a[r]-=orz;
 82                 if(a[l]&1)a[l]+=orz;
 83                 if(a[r]&1)a[r]+=orz;
 84                 a[l]>>=1,a[r]>>=1;
 85             }
 86         }
 87     }
 88 }
 89 int main(){
 90     scanf("%d",&m);
 91     for(int i=0;i<m;++i)scanf("%d",a+i),_o.a[i]=_a.a[i]=_x.a[i]=a[i];
 92     for(int i=0;i<m;++i)scanf("%d",b+i),_o.b[i]=_a.b[i]=_x.b[i]=b[i];
 93     n=1;
 94     while(n<m)n<<=1;
 95     fwt_or(_o.a),fwt_or(_o.b),fwt_and(_a.a),fwt_and(_a.b),fwt_xor(_x.a),fwt_xor(_x.b);
 96     _o.mul(),_a.mul(),_x.mul();
 97     ifwt_or(_o.c),ifwt_and(_a.c),ifwt_xor(_x.c);
 98     _o.out(),_a.out(),_x.out();
 99     return 0;
100 }
    View Code

    后缀自动机

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int n,a[5000005];
 8 int ch[10000005][26],nx[10000005],mx[10000005],la=1,_=1;
 9 ll res,ans;
10 int main(){
11     char in=getchar();
12     while(in>96)a[++n]=in-97,in=getchar();
13     for(int i=1;i<=n;++i){
14         int x=++_,y=la,c=a[i];
15         mx[x]=mx[y]+1;
16         for(;y&&!ch[y][c];y=nx[y])ch[y][c]=x;
17         if(!y)nx[x]=1,res+=mx[x];
18         else if(mx[ch[y][c]]==mx[y]+1)nx[x]=ch[y][c],res+=mx[x]-mx[nx[x]];
19         else{
20             int p=++_,o=ch[y][c];
21             mx[p]=mx[y]+1;
22             nx[p]=nx[o],nx[o]=nx[x]=p;
23             res+=mx[x]-mx[p];
24             memcpy(ch[p],ch[o],sizeof(ch[o]));
25             for(;y&&ch[y][c]==o;y=nx[y])ch[y][c]=p;
26         }
27         la=x,ans^=res;
28     }
29     printf("%lld
",ans);
30     return 0;
31 }
    View Code

    回文自动机

     1 #include<bits/stdc++.h>
 2 #define ll   long long
 3 #define pb   push_back
 4 #define mp   make_pair
 5 #define orz  1000000007
 6 using namespace std;
 7 int n,a[300005],len[300005],cnt[300005],ch[300005][26],_,la,nx[300005];
 8 ll ans;
 9 int ne(int l){
10     len[_]=l,cnt[_]=0;
11     return _++;
12 }
13 int get(int x,int y){
14     while(a[y-len[x]-1]!=a[y])x=nx[x];
15     return x;
16 }
17 int main(){
18     char in=getchar();
19     while(in>96)a[++n]=in-97,in=getchar();
20     a[0]=-1,ne(0),ne(-1),la=0,nx[0]=1;
21     for(int i=1;i<=n;++i){
22         int o=get(la,i),k=a[i];
23         if(!ch[o][k]){
24             int p=ne(len[o]+2);
25             nx[p]=ch[get(nx[o],i)][k];
26             ch[o][k]=p;
27         }
28         ++cnt[la=ch[o][k]];
29     }
30     for(int i=_-1;i>1;--i)cnt[nx[i]]+=cnt[i],ans=max(ans,len[i]*1ll*cnt[i]);
31     printf("%lld
",ans);
32     return 0;
33 }
    View Code
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  • 原文地址：https://www.cnblogs.com/skylineidolon/p/9134026.html
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