poj 2987 最大闭合子图

思路：

这题考的是最大闭权图。只要知道怎么求最大闭权图就知道怎么做。但好像有点卡模版，要高效的模版才行。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define Maxn 6010
#define Maxm 200000
#define LL __int64
#define Abs(a) (a)>0?(a):(-a)
using namespace std;
struct Edge{
    int from,to,next;
    LL val;
}edge[Maxm];
LL value[Maxn],inf=0;
int index[Maxn],work[Maxn],dis[Maxn],q[Maxn],e,vi[Maxn];
inline void addedge(int from,int to,LL val)//有向边
{
    edge[e].from=from;
    edge[e].to=to;
    edge[e].val=val;
    edge[e].next=index[from];
    index[from]=e++;
    edge[e].from=to;
    edge[e].to=from;
    edge[e].val=0;
    edge[e].next=index[to];
    index[to]=e++;
}
void init()
{
    e=0;
    memset(index,-1,sizeof(index));
    memset(vi,0,sizeof(vi));
    inf=0;
}
void add(int u,int v,LL c)
{
    edge[e].to=v;edge[e].val=c;edge[e].next=index[u];index[u]=e++;
    edge[e].to=u;edge[e].val=0;edge[e].next=index[v];index[v]=e++;
}
void DFS(int u)
{
    vi[u]=1;
    int i,v;
    for(i=index[u];i!=-1;i=edge[i].next)
        if(edge[i].val&&!vi[edge[i].to])
            DFS(edge[i].to);
}
int bfs(int S,int T)
{
    int rear=0;
    memset(dis,-1,sizeof(dis));
    dis[S]=0;q[rear++]=S;
    for(int i=0;i<rear;i++)
    {
        for(int j=index[q[i]];j!=-1;j=edge[j].next)
        {
            if(edge[j].val&&dis[edge[j].to]==-1)
            {
                dis[edge[j].to]=dis[q[i]]+1;
                q[rear++]=edge[j].to;
                if(edge[j].to==T) return 1;
            }
        }
    }
    return 0;
}
LL dfs(int cur,LL a,int T)
{
    if(cur==T) return a;
    for(int &i=work[cur];i!=-1;i=edge[i].next)
    {
        if(edge[i].val&&dis[edge[i].to]==dis[cur]+1)
        {
            int t=dfs(edge[i].to,min(a,edge[i].val),T);
            if(t)
            {
                edge[i].val-=t;
                edge[i^1].val+=t;
                return t;
            }
        }
    }
    return 0;
}
LL Dinic(int S,int T)
{
    LL ans=0;
    while(bfs(S,T))
    {
        memcpy(work,index,sizeof(index));
        while(int t=dfs(S,inf,T)) ans+=t;
    }
    return ans;
}

int main()
{
    int n,m,i,a,b;
    LL sum=0;
    scanf("%d%d",&n,&m);
    init();
    for(i=1;i<=n;i++)
    {
        scanf("%I64d",value+i);
        if(value[i]>0)
        {
            sum+=value[i];
            addedge(0,i,value[i]);
        }
        else
        addedge(i,n+1,-value[i]);
        inf+=Abs(value[i]);
    }
    inf++;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        addedge(a,b,inf);
    }
    LL ans=Dinic(0,n+1);//起始点和结束点
    //DFS(0);//寻找S集合
    int num=0;
    for(i=1;i<=n;i++)
        if(dis[i]>=0)
            num++;
    printf("%d %I64d
",num,sum-ans);
    return 0;
}