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传说中的可持久化线段树优化网络流。。。
做了一些预备题后最小割建图还是比较简单,但边数是n^2。
每次可以通过向值域中的区间连边,使边数降为nlogn,要满足j<i,打一个可持久化,每次搞完再insert即可
蒯一个PoPoQQQ大爷的图,很清楚啊。。。
// MADE BY QT666 #include<cstdio> #include<algorithm> #include<cmath> #include<iostream> #include<cstring> #define RG register using namespace std; typedef long long ll; const int N=1000050; const int Inf=19260817; int gi(){ int x=0,flag=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-') flag=-1;ch=getchar();} while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x*flag; } int head[N],nxt[N],to[N],s[N],cnt=1,level[N],q[N],S,T,F; int a[5050],b[5050],w[5050],l[5050],r[5050],p[5050],n,tot; int sz,ls[100050],rs[100050],root[100050],hsh[100050],sum,goal,hh=0; inline void Addedge(RG int x,RG int y,RG int z) { to[++cnt]=y,s[cnt]=z,nxt[cnt]=head[x],head[x]=cnt; } inline void lnk(RG int x,RG int y,RG int z){ if(!x||!y) return; Addedge(x,y,z),Addedge(y,x,0); } inline bool bfs(){ for(RG int i=1;i<=sz;i++) level[i]=0; q[0]=S,level[S]=1;int t=0,sum=1; while(t<sum){ int x=q[t++]; if(x==T) return 1; for(RG int i=head[x];i;i=nxt[i]){ int y=to[i]; if(s[i]&&level[y]==0){ level[y]=level[x]+1; q[sum++]=y; } } } return 0; } inline int dfs(RG int x,int maxf){ if(x==T) return maxf; int ret=0; for(RG int i=head[x];i;i=nxt[i]){ int y=to[i],f=s[i]; if(level[y]==level[x]+1&&f){ int minn=min(f,maxf-ret); f=dfs(y,minn); s[i]-=f,s[i^1]+=f,ret+=f; if(ret==maxf) break; } } if(!ret) level[x]=0; return ret; } inline void Dinic(){ while(bfs()) F+=dfs(S,Inf); } inline void insert(int l,int r,int x,int &y,int v){ y=++sz;ls[y]=ls[x],rs[y]=rs[x];hh++; lnk(x,y,Inf);lnk(goal,y,Inf); if(l==r) return; int mid=(l+r)>>1; if(v<=mid) insert(l,mid,ls[x],ls[y],v); else insert(mid+1,r,rs[x],rs[y],v); } inline void query(int x,int L,int R,int xl,int xr){ if(!x) return; if(xl<=L&&R<=xr){lnk(x,goal,Inf);return;} int mid=(L+R)>>1; if(xl<=mid) query(ls[x],L,mid,xl,xr); if(xr>mid) query(rs[x],mid+1,R,xl,xr); } int main(){ n=gi();S=2*n+1;T=2*n+2;sz=T; for(RG int i=1;i<=n;i++){ a[i]=gi(),b[i]=gi(),w[i]=gi(),l[i]=gi(),r[i]=gi(),p[i]=gi(); tot+=b[i],tot+=w[i];hsh[++sum]=a[i]; } sort(hsh+1,hsh+1+sum);sum=unique(hsh+1,hsh+sum+1)-hsh-1; for(int i=1;i<=n;i++){ a[i]=lower_bound(hsh+1,hsh+1+sum,a[i])-hsh; l[i]=lower_bound(hsh+1,hsh+1+sum,l[i])-hsh; r[i]=upper_bound(hsh+1,hsh+1+sum,r[i])-hsh-1; lnk(S,i,w[i]);lnk(i,T,b[i]);lnk(i+n,i,p[i]); } for(int i=1;i<=n;i++){ goal=i+n;if(l[i]<=r[i]) query(root[i-1],1,sum,l[i],r[i]); goal=i;insert(1,sum,root[i-1],root[i],a[i]); } Dinic();printf("%d ",tot-F); return 0; }