hdu5820
官方题解：http://www.cnblogs.com/duoxiao/p/5777700.html
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 struct node{int l,r,s;} tr[500010*40]; 5 vector<int> g[50010]; 6 int pre[50010],h[50010],n,t; 7 int build(int l,int r) 8 { 9 ++t; tr[t].s=0; 10 if (l==r) return t; 11 int m=(l+r)>>1,q=t; 12 tr[t].l=build(l,m); 13 tr[t].r=build(m+1,r); 14 return q; 15 } 16 17 int ask(int i,int l,int r,int x,int y) 18 { 19 if (x>y) return 0; 20 if (x<=l&&y>=r) return tr[i].s; 21 else { 22 int m=(l+r)>>1,s=0; 23 if (x<=m) s+=ask(tr[i].l,l,m,x,y); 24 if (y>m) s+=ask(tr[i].r,m+1,r,x,y); 25 return s; 26 } 27 } 28 29 int add(int i,int l,int r,int x) 30 { 31 ++t; 32 if (l==r) 33 { 34 tr[t].s=tr[i].s+1; 35 return t; 36 } 37 int m=(l+r)>>1,q=t; 38 if (x<=m) 39 { 40 tr[q].r=tr[i].r; 41 tr[q].l=add(tr[i].l,l,m,x); 42 } 43 else { 44 tr[q].l=tr[i].l; 45 tr[q].r=add(tr[i].r,m+1,r,x); 46 } 47 tr[q].s=tr[tr[q].l].s+tr[tr[q].r].s; 48 return q; 49 } 50 51 52 bool check() 53 { 54 memset(pre,0,sizeof(pre)); 55 for (int i=1; i<=50000; i++) 56 { 57 h[i]=h[i-1]; 58 for (int j=0; j<g[i].size(); j++) 59 { 60 int l=(!j)?0:g[i][j-1]; 61 int r=(j==g[i].size()-1)?50001:g[i][j+1]; 62 // cout <<l<<" "<<r<<" "<<pre[g[i][j]]<<" "<<ask(h[i],1,50000,l+1,r-1)<<endl; 63 if (ask(h[i],1,50000,l+1,r-1)!=ask(h[pre[g[i][j]]],1,50000,l+1,r-1)) return 0; 64 } 65 for (int j=0; j<g[i].size(); j++) 66 { 67 h[i]=add(h[i],1,50000,g[i][j]); 68 pre[g[i][j]]=i; 69 } 70 } 71 return 1; 72 } 73 74 int main() 75 { 76 scanf("%d",&n); 77 t=0; 78 h[0]=build(1,50000); int tmp=t; 79 while (n) 80 { 81 t=tmp; 82 for (int i=1; i<=n; i++) 83 { 84 int x,y; 85 scanf("%d%d",&x,&y); 86 g[x].push_back(y); 87 } 88 for (int i=1; i<=50000; i++) 89 { 90 sort(g[i].begin(),g[i].end()); 91 g[i].erase(unique(g[i].begin(),g[i].end()),g[i].end()); 92 } 93 if (check()) puts("YES"); else puts("NO"); 94 for (int i=1; i<=50000; i++) g[i].clear(); 95 scanf("%d",&n); 96 } 97 }
hdu5788
官方题解：http://www.cnblogs.com/duoxiao/p/5777661.html
其实有个简单的方法求子树的[t/2]名次：直接在dfs序上建立可持久化线段树查询即可
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 struct way{int po,next;} e[100010]; 6 struct node{int s,l,r;} tr[100010*40]; 7 int p[100010],l[100010],h[100010],w[100010],cur[100010],a[100010],r[100010],len,n,tot,t; 8 ll c[100010],ans; 9 void add(int x,int y) 10 { 11 e[++len].po=y; 12 e[len].next=p[x]; 13 p[x]=len; 14 } 15 16 int build(int l,int r) 17 { 18 ++t; tr[t].s=0; 19 if (l==r) return t; 20 int m=(l+r)>>1,q=t; 21 tr[t].l=build(l,m); 22 tr[t].r=build(m+1,r); 23 return q; 24 } 25 26 int mid(int a,int b,int l,int r,int k) 27 { 28 if (l==r) return l; 29 else { 30 int m=(l+r)>>1,s=tr[tr[b].l].s-tr[tr[a].l].s; 31 if (k<=s) return mid(tr[a].l,tr[b].l,l,m,k); 32 else return mid(tr[a].r,tr[b].r,m+1,r,k-s); 33 } 34 } 35 36 int add(int i,int l,int r,int x) 37 { 38 ++t; 39 if (l==r) 40 { 41 tr[t].s=tr[i].s+1; 42 return t; 43 } 44 int m=(l+r)>>1,q=t; 45 if (x<=m) 46 { 47 tr[q].r=tr[i].r; 48 tr[q].l=add(tr[i].l,l,m,x); 49 } 50 else { 51 tr[q].l=tr[i].l; 52 tr[q].r=add(tr[i].r,m+1,r,x); 53 } 54 tr[q].s=tr[tr[q].l].s+tr[tr[q].r].s; 55 return q; 56 } 57 58 void dfs1(int x) 59 { 60 l[x]=++tot; h[tot]=add(h[tot-1],1,100000,a[x]); 61 for (int i=p[x]; i; i=e[i].next) 62 { 63 int y=e[i].po; 64 dfs1(y); 65 } 66 w[x]=mid(h[l[x]-1],h[tot],1,100000,(tot-l[x]+2)/2); 67 if (l[x]==tot) cur[x]=0; else cur[x]=mid(h[l[x]-1],h[tot],1,100000,(tot-l[x]+2)/2+1)-w[x]; 68 } 69 70 void work(int x,int w) 71 { 72 for (int i=x; i; i-=i&(-i)) c[i]+=w; 73 } 74 75 ll ask(int x) 76 { 77 ll s=0; 78 for (int i=x; i<=100000; i+=i&(-i)) s+=c[i]; 79 return s; 80 } 81 82 void dfs2(int x) 83 { 84 ll wh=(p[x]==0)?100000-a[x]:cur[x]-w[x]; 85 ans=max(ans,ask(a[x])+wh); 86 if (cur[x]) work(w[x],cur[x]); 87 for (int i=p[x]; i; i=e[i].next) 88 { 89 int y=e[i].po; 90 dfs2(y); 91 } 92 if (cur[x]) work(w[x],-cur[x]); 93 } 94 95 int main() 96 { 97 t=0; h[0]=build(1,100000); 98 int tmp=t; 99 while (scanf("%d",&n)!=EOF) 100 { 101 t=tmp; 102 len=0; memset(p,0,sizeof(p)); 103 for (int i=1; i<=n; i++) scanf("%d",&a[i]); 104 for (int i=2; i<=n; i++) 105 { 106 int fa; 107 scanf("%d",&fa); 108 add(fa,i); 109 } 110 tot=0; h[1]=h[0]; dfs1(1); 111 ll s=0; 112 for (int i=1; i<=n; i++) s+=w[i]; 113 // for (int i=1; i<=n; i++) cout <<"*"<<w[i]<<" "<<cur[i]<<endl; 114 memset(c,0,sizeof(c)); ans=0; 115 dfs2(1); 116 printf("%lld ",ans+s); 117 } 118 }
hdu3605
把n按照可选择方案二进制状态压缩，建图最大流即可
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 const int inf=1e9+7; 5 struct way{int po,flow,next;} e[2400010]; 6 int t,n,p[2010],pre[2010],numh[2010],cur[2010],h[2010],d[2010],b[2010],m,len; 7 8 void add(int x,int y,int f) 9 { 10 e[++len].po=y; 11 e[len].flow=f; 12 e[len].next=p[x]; 13 p[x]=len; 14 } 15 void build(int x, int y, int f) 16 { 17 add(x,y,f); 18 add(y,x,0); 19 } 20 int sap() 21 { 22 memset(numh,0,sizeof(numh)); 23 memset(h,0,sizeof(h)); 24 numh[0]=t+1; 25 for (int i=0; i<=t; i++) cur[i]=p[i]; 26 int j,u=0,s=0,neck=inf; 27 while (h[0]<t+1) 28 { 29 d[u]=neck; 30 bool ch=1; 31 for (int i=cur[u]; i!=-1; i=e[i].next) 32 { 33 j=e[i].po; 34 if (e[i].flow>0&&h[u]==h[j]+1) 35 { 36 neck=min(neck,e[i].flow); 37 cur[u]=i; 38 pre[j]=u; u=j; 39 if (u==t) 40 { 41 s+=neck; 42 while (u) 43 { 44 u=pre[u]; 45 j=cur[u]; 46 e[j].flow-=neck; 47 e[j^1].flow+=neck; 48 } 49 neck=inf; 50 } 51 ch=0; 52 break; 53 } 54 } 55 if (ch) 56 { 57 if (--numh[h[u]]==0) return s; 58 int q=-1,tmp=t; 59 for (int i=p[u]; i!=-1; i=e[i].next) 60 { 61 j=e[i].po; 62 if (e[i].flow&&h[j]<tmp) 63 { 64 tmp=h[j]; 65 q=i; 66 } 67 } 68 cur[u]=q; h[u]=tmp+1; 69 numh[h[u]]++; 70 if (u) 71 { 72 u=pre[u]; 73 neck=d[u]; 74 } 75 } 76 } 77 return s; 78 } 79 80 int main() 81 { 82 while (scanf("%d%d",&n,&m)!=EOF) 83 { 84 len=-1; 85 memset(p,255,sizeof(p)); 86 memset(b,0,sizeof(b)); 87 for (int i=1; i<=n; i++) 88 { 89 int st=0,x; 90 for (int j=0; j<m; j++) 91 { 92 scanf("%d",&x); 93 st|=(x&1)<<j; 94 } 95 b[st]++; 96 } 97 t=(1<<m)+m+1; 98 for (int i=0; i<(1<<m); i++) 99 { 100 for (int j=0; j<m; j++) 101 if ((i>>j)&1) build(i+m+1,j+1,b[i]); 102 build(0,i+m+1,b[i]); 103 } 104 for (int i=1; i<=m; i++) 105 { 106 int x; 107 scanf("%d",&x); 108 build(i,t,x); 109 } 110 if (sap()==n) puts("YES"); else puts("NO"); 111 } 112 }
hdu5988
费用取对数，简单的费用流
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 using namespace std; 6 struct way{int po,next,flow; double cost;} e[200010]; 7 const int inf=100000007; 8 const double eps=1e-7; 9 int pre[110],p[110],cur[110],q[4000010]; 10 double d[110]; 11 bool v[110]; 12 int len,n,m,t,k; 13 14 void add(int x,int y,int f,double c) 15 { 16 e[++len].po=y; 17 e[len].flow=f; 18 e[len].cost=c; 19 e[len].next=p[x]; 20 p[x]=len; 21 } 22 23 void build(int x,int y, int f, double c) 24 { 25 add(x,y,f,c); 26 add(y,x,0,-c); 27 } 28 29 bool spfa() 30 { 31 int f=1,r=1; 32 for (int i=1; i<=t; i++) d[i]=inf; 33 memset(v,0,sizeof(v)); 34 d[0]=0; q[1]=0; 35 while (f<=r) 36 { 37 int x=q[f++]; 38 v[x]=0; 39 for (int i=p[x]; i!=-1; i=e[i].next) 40 { 41 int y=e[i].po; 42 if (e[i].flow&&d[x]+e[i].cost+eps<d[y]) 43 { 44 d[y]=d[x]+e[i].cost; 45 pre[y]=x; cur[y]=i; 46 if (!v[y]) 47 { 48 q[++r]=y; 49 v[y]=1; 50 } 51 } 52 } 53 } 54 return d[t]<inf; 55 } 56 57 double mincost() 58 { 59 int j; 60 double s=0; 61 while (spfa()) 62 { 63 int neck=inf; 64 for (int i=t; i; i=pre[i]) 65 { 66 j=cur[i]; 67 neck=min(neck,e[j].flow); 68 } 69 s+=d[t]*neck; 70 for (int i=t; i; i=pre[i]) 71 { 72 j=cur[i]; 73 e[j].flow-=neck; 74 e[j^1].flow+=neck; 75 } 76 } 77 return s; 78 } 79 80 int main() 81 { 82 int cas; 83 scanf("%d",&cas); 84 while (cas--) 85 { 86 scanf("%d%d",&n,&m); 87 t=n+1; len=-1; memset(p,255,sizeof(p)); 88 for (int i=1; i<=n; i++) 89 { 90 int x,y; 91 scanf("%d%d",&x,&y); 92 if (x-y>0) build(0,i,x-y,0); 93 else if (x-y<0) build(i,t,y-x,0); 94 } 95 for (int i=1; i<=m; i++) 96 { 97 int x,y,f; double c; 98 scanf("%d%d%d%lf",&x,&y,&f,&c); 99 c=-log2(1.0-c); 100 if (f) build(x,y,1,0); 101 if (f>1) build(x,y,f-1,c); 102 } 103 double ans=mincost(); 104 ans=pow(2,-ans); 105 printf("%.2lf ",1.0-ans); 106 } 107 }
hdu5985
无穷项一般要么高斯消元要么取有限项保留对应精度即可，这里只要求出h[i][k]表示第i个在第k次操作后被全部移走的概率就可以轻易计算出答案
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 double h[15][80],ans[15],p[15]; 5 int a[15],n; 6 int main() 7 { 8 int cas; 9 scanf("%d",&cas); 10 while (cas--) 11 { 12 scanf("%d",&n); 13 for (int i=1; i<=n; i++) 14 scanf("%d%lf",&a[i],&p[i]); 15 if (n==1) 16 { 17 printf("%.6lf ",1.0); 18 continue; 19 } 20 for (int i=1; i<=n; i++) 21 { 22 double tmp=1.0; 23 for (int j=1; j<=76; j++) 24 { 25 tmp*=p[i]; 26 h[i][j]=pow(1-tmp,a[i]); 27 } 28 } 29 for (int i=1; i<=n; i++) 30 { 31 ans[i]=0; 32 for (int k=1; k<=75; k++) 33 { 34 double tmp=1.0; 35 for (int j=1; j<=n; j++) 36 if (j!=i) tmp*=h[j][k]; 37 ans[i]+=(h[i][k+1]-h[i][k])*tmp; 38 } 39 } 40 for (int i=1; i<=n; i++) 41 { 42 printf("%.6lf",ans[i]); 43 if (i==n) puts(""); else printf(" "); 44 } 45 } 46 }