POJ 2987 Firing 网络流 最大权闭合图

http://poj.org/problem?id=2987

https://blog.csdn.net/u014686462/article/details/48533253

给一个闭合图，要求输出其最大权闭合图的权值和需要选的最少点数，最大权闭合图定义和网络流连边方式见博客。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<iostream>
 #include<queue>
 using namespace std;
 #define LL long long
 const int maxn=;
 const LL minf=(LL)1e14;
 int n,m,s,t;
 LL val[maxn]={};
 struct nod{
     int y,next;LL v;
 }e[maxn*];
 int head[maxn],tot=;
 queue<int>q;
 int dep[maxn]={},vis[maxn]={},id[maxn]={},tai=;
 void init(int x,int y,LL v){
     e[++tot].y=y;e[tot].v=v;e[tot].next=head[x];head[x]=tot;
 }
 bool dfs(){
     memset(dep,,sizeof(dep));
     q.push(s);dep[s]=;
     while(!q.empty()){
         int x=q.front();q.pop();
         for(int i=head[x];i;i=e[i].next){
             if(e[i].v&&!dep[e[i].y]){
                 dep[e[i].y]=dep[x]+;
                 q.push(e[i].y);
             }
         }
     }
     return dep[t];
 }
 LL dfs1(int x,LL fc){
     if(x==t){
         return fc;
     }
     LL he=,z;
     for(int i=head[x];i;i=e[i].next){
         if(dep[x]+==dep[e[i].y]){
             z=dfs1(e[i].y,min(fc-he,e[i].v));
             he+=z;e[i].v-=z;e[i^].v+=z;
             if(he==fc)break;
         }
     }
     return he;
 }
 void dfs2(int x){
     if(x==t)return;
     if(x!=s)id[++tai]=x;
     vis[x]=;
     for(int i=head[x];i;i=e[i].next){
         if(e[i].v&&!vis[e[i].y]){
             dfs2(e[i].y);
         }
     }
 }
 int main(){
     int x,y; LL ans=;
     scanf("%d%d",&n,&m);
     s=n+;t=s+;
     for(int i=;i<=n;i++){
         scanf("%lld",&val[i]);
         if(val[i]>=){init(s,i,val[i]);init(i,s,);ans=ans+val[i];}
         else {init(i,t,-val[i]);init(t,i,);}
     }
     for(int i=;i<=m;i++){
         scanf("%d%d",&x,&y);
         init(x,y,minf);init(y,x,);
     }
     while(dfs())ans-=dfs1(s,minf);
     dfs2(s);
     printf("%d ",tai);
     printf("%lld\n",ans);
     return ;
 }