BZOJ 1497 最大获利(最大权闭合子图)

http://www.lydsy.com/JudgeOnline/problem.php?id=1497

思路:由题意可以得知，每个顾客都依赖2个中转站，那么让中转站连有向边到汇点，流量为它的建设费用，源点连到每个顾客，流量为赚的钱，然后每个顾客到它依赖的中转站连流量为inf的边

 #include<algorithm>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iostream>
 #define inf 0x7fffffff
 int tot,go[],first[],next[],flow[];
 int op[],S,T,nodes,dis[],cnt[],n,m;
 int read(){
     int t=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (''<=ch&&ch<=''){t=t*+ch-'';ch=getchar();}
     return t*f;
 }
 void insert(int x,int y,int z){
     tot++;
     go[tot]=y;
     next[tot]=first[x];
     first[x]=tot;
     flow[tot]=z;
 }
 void add(int x,int y,int z){
     insert(x,y,z);op[tot]=tot+;
     insert(y,x,);op[tot]=tot-;
 }
 int dfs(int x,int f){
     if (x==T) return f;
     int mn=nodes,sum=;
     for (int i=first[x];i;i=next[i]){
         int pur=go[i];
         if (flow[i]&&dis[pur]+==dis[x]){
             int save=dfs(pur,std::min(f-sum,flow[i]));
             sum+=save;
             flow[i]-=save;
             flow[op[i]]+=save;
             if (f==sum||dis[S]>=nodes) return sum;
         }
         if (flow[i]) mn=std::min(mn,dis[pur]);
     }
     if (sum==){
         cnt[dis[x]]--;
         if (cnt[dis[x]]==){
             dis[S]=nodes;
         }
         else{
             dis[x]=mn+;
             cnt[dis[x]]++;
         }
     }
     return sum;
 }
 int main(){
     n=read(),m=read();
     nodes=n+m+;
     T=n+m+;S=;
     for (int i=;i<=n;i++){
      int x=read();
      add(m+i,T,x);
     }
     int sum=;
     for (int i=;i<=m;i++){
         int x=read(),y=read(),z=read();
         add(S,i,z);
         add(i,x+m,inf);
         add(i,y+m,inf);
         sum+=z;
     }
     int ans=;
     while (dis[S]<nodes) ans+=dfs(S,inf);
     printf("%d\n",std::max(sum-ans,));
 }