hdu 3491 Thieves

权值是在点上的，进行拆点，入边为i,出边为i+n,容量为该点权值,注意S，H两点不会算在最小割中，所以要么这两点不拆点，要么拆点后容量为无穷,其他m*2条边的容量都为无穷

 

一开始一直wa，想当然的加入两条边(i+n,j,inf)和(j,i+n,inf),其实应该是(i+n,j,inf)和(j+n,i,inf),出边应该是i+n而不是i!!!

 

代码：

#include<iostream>#include<memory.h>#include<string>#include<cstdio>#include<algorithm>#include<math.h>#include<stack>#include<queue>using namespace std; const int MAX=205;const int inf=999999;struct node{int v,c,next;}g[MAX*MAX];int adj[MAX],dis[MAX],cur[MAX],num[MAX],pre[MAX];int n,m,e,s,t,vn,S,H;void add(int u,int v,int c) { g[e].v=v; g[e].c=c; g[e].next=adj[u]; adj[u]=e++; g[e].v=u; g[e].c=0; g[e].next=adj[v]; adj[v]=e++; } int sap() { int i,u,v,flag,aug=inf+1,flow=0; //memset(pre,0,sizeof(pre)); for(i=0;i<=vn;i++) { cur[i]=adj[i]; num[i]=dis[i]=0; } num[0]=vn; pre[s]=u=s; while(dis[s]<vn) { flag=0; for(i=adj[u];i!=-1;i=g[i].next) { v=g[i].v; if(g[i].c&&dis[u]==dis[v]+1) { flag=1; aug=min(aug,g[i].c); pre[v]=u; cur[u]=i; u=v; if(u==t) { flow+=aug; while(u!=s) { u=pre[u]; g[cur[u]].c-=aug; g[cur[u]^1].c+=aug; } aug=inf+1; } break; } } if(flag) continue; if(--num[dis[u]]==0) break; for(dis[u]=inf,i=adj[u];i!=-1;i=g[i].next) { v=g[i].v; if(g[i].c&&dis[v]<dis[u]) { dis[u]=dis[v]; cur[u]=i; } } dis[u]++; num[dis[u]]++; u=pre[u]; } return flow; } int main(){int i,j,k,T;scanf("%d",&T);while(T--){scanf("%d%d%d%d",&n,&m,&S,&H);e=0;vn=2*n+2;s=S+n;t=H;memset(adj,-1,sizeof(adj));add(s,S,inf);add(H+n,T,inf);for(i=1;i<=n;i++){scanf("%d",&k);if(i!=S&&i!=H){add(i,i+n,k);//add(i+n,i,k);}else{add(i,i+n,inf);//add(i+n,i,inf);}}while(m--){scanf("%d%d",&i,&j);if(i!=s&&j!=t){add(i+n,j,inf);add(j+n,i,inf);//不是add(j,i+n)!!!,i+n为出边，i为入边}else{if(i==s){add(s,j,inf);//add(j,s,inf);}else{add(i+n,t,inf);//add(t,i+n,inf);}}}printf("%d/n",sap());}return 0;}