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=10010;
const LL minf=(LL)1e14;
int n,m,s,t;
LL val[maxn]={};
struct nod{
    int y,next;LL v;
}e[maxn*20];
int head[maxn],tot=1;
queue<int>q;
int dep[maxn]={},vis[maxn]={},id[maxn]={},tai=0;
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,0,sizeof(dep));
    q.push(s);dep[s]=1;
    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]+1;
                q.push(e[i].y);
            }
        }
    }
    return dep[t];
}
LL dfs1(int x,LL fc){
    if(x==t){
        return fc;
    }
    LL he=0,z;
    for(int i=head[x];i;i=e[i].next){
        if(dep[x]+1==dep[e[i].y]){
            z=dfs1(e[i].y,min(fc-he,e[i].v));
            he+=z;e[i].v-=z;e[i^1].v+=z;
            if(he==fc)break;
        }
    }
    return he;
}
void dfs2(int x){
    if(x==t)return;
    if(x!=s)id[++tai]=x;
    vis[x]=1;
    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=0;
    scanf("%d%d",&n,&m);
    s=n+1;t=s+1;
    for(int i=1;i<=n;i++){
        scanf("%lld",&val[i]);
        if(val[i]>=0){init(s,i,val[i]);init(i,s,0);ans=ans+val[i];}
        else {init(i,t,-val[i]);init(t,i,0);}
    }
    for(int i=1;i<=m;i++){
        scanf("%d%d",&x,&y);
        init(x,y,minf);init(y,x,0);
    }
    while(dfs())ans-=dfs1(s,minf);
    dfs2(s);
    printf("%d ",tai);
    printf("%lld
",ans);
    return 0;
}