poj1815Friendship(最小割求割边）

链接

题意为去掉多少个顶点使图不连通，求顶点连通度问题。拆点，构造图，对于<u,v>可以变成<u2,v1> <v2,u1>容量为无穷，<u1,u2>容量为1.那么求出来的最大流（即最小割）就为所需要删除的顶点个数，需要字典序输出，从小到大枚举顶点，如果不加入当前点，最小割变小了的话 ，说明这个点是肯定要删除的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f
const int N = 415;
#define M 160015
struct node
{
    int u,v,next;
    int w;
} edge[M<<1];
int head[N],t,vis[N],pp[N],dis[N];
int o[N];
int st,en;
int x[N][N],f[N];
void init()
{
    t=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
    edge[t].u = u;
    edge[t].v = v;
    edge[t].w = w;
    edge[t].next = head[u];
    head[u] = t++;
    edge[t].u = v;
    edge[t].v = u;
    edge[t].w = 0;
    edge[t].next = head[v];
    head[v] = t++;
}
int bfs()
{
    int i,u;
    int w;
    memset(dis,-1,sizeof(dis));
    queue<int>q;
    q.push(st);
    dis[st] = 0;
    while(!q.empty())
    {
        u = q.front();
        q.pop();
        for(i = head[u] ; i != -1 ; i = edge[i].next)
        {
            int v = edge[i].v;
            w = edge[i].w;
            if(dis[v]<0&&w>0)
            {
                dis[v] = dis[u]+1;
                q.push(v);
            }
        }
    }
    if(dis[en]>0) return 1;
    return 0;
}
int dfs(int u,int te)
{
    int i;
    int s;
    if(u==en) return te;
    for(i = head[u] ; i != -1 ; i = edge[i].next)
    {
        int v = edge[i].v;
        int w = edge[i].w;
        if(w>0&&dis[v]==dis[u]+1&&(s=dfs(v,min(te,w))))
        {
            edge[i].w-=s;
            edge[i^1].w+=s;
            return s;
        }
    }
    dis[u] = -1;
    return 0;
}
int dinic()
{
    int flow = 0;
    int res;
    while(bfs())
    {
        while(res = dfs(st,INF))
        flow+=res;
    }
    return flow;
}
int main()
{
    int n,i,j;
    while(scanf("%d%d%d",&n,&st,&en)!=EOF)
    {
        init();
        // memset(x,0,sizeof(x));
        memset(f,0,sizeof(f));
        st+=n;
        for(i = 1; i <= n ; i++)
        {
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&x[i][j]);
                if(i==j)
                {
                    add(i,i+n,1);
                }
                else if(x[i][j])
                {
                    add(i+n,j,INF);
                }
            }
        }
        if(x[st-n][en])
        {
            puts("NO ANSWER!");
            continue;
        }
        int ans = dinic();
        int cnt = 0;
        for(i = 1; i <= n ; i++)
        {
            if(ans==0) break;
            if(i==st-n||i==en) continue;
            f[i] = 1;
            init();
            for(j = 1; j <= n ; j++)
            {
                if(f[j]) continue;
                for(int e = 1; e <= n ; e++)
                {
                    if(f[e]) continue;
                    if(j==e)
                        add(j,j+n,1);
                    else if(x[j][e])
                    {
                        add(j+n,e,INF);
                    }
                }
            }
            int ts = dinic();
            if(ts<ans)
            {
                cnt++;
                ans = ts;
            }
            else f[i] = 0;
        }
        cout<<cnt<<endl;
        for(j = 1; j <= n; j++)
            if(f[j])
                printf("%d ",j);
        printf("
");
    }
    return 0;
}