USACO5.4-TeleCowmunication

题目大意:给出一个无向图,要求删除尽量少的点,使给定的2点间不再连通,并输出字典序最小的方案
题型:图论-网络流
此题难点在于建图,后面就是套网络流的模板.
将点看成边,例如第i个点可以看成一条有向边<i*2-1,i*2>,容量为1.
如果j点和i点邻接,那么新建2条容量为无穷大的有向边<i*2,j*2-1>,<j*2,i*2-1>.
然后应用最大流最小割定理,求最大流即为第一问答案.
接着枚举删除每一个点i(即删除有向边),看最大流是否减少1,如果是则该点在最小割中,然后真的把这一点删点.
枚举完每个点之后别忘了将残量网络还原.
至于为什么要这样建图, 一时间说不清楚......

Executing...
Test 1: TEST OK [0.008 secs, 3852 KB]
Test 2: TEST OK [0.014 secs, 3852 KB]
Test 3: TEST OK [0.005 secs, 3852 KB]
Test 4: TEST OK [0.022 secs, 3852 KB]
Test 5: TEST OK [0.011 secs, 3852 KB]
Test 6: TEST OK [0.019 secs, 3852 KB]
Test 7: TEST OK [0.019 secs, 3852 KB]
Test 8: TEST OK [0.014 secs, 3852 KB]
Test 9: TEST OK [0.032 secs, 3852 KB]
Test 10: TEST OK [0.046 secs, 3852 KB]
Test 11: TEST OK [0.068 secs, 3852 KB]

All tests OK.
Your program ('telecow') produced all correct answers! This is your
submission #3 for this problem. Congratulations!

#include <iostream>
#include <cstring>
#include <queue>
#include <stdio.h>
#define msize 210
#define INF 1000000000
using namespace std;

int origin[msize][msize]={0};
int r[msize][msize]={0}; //残留网络，初始为原图
bool visited[msize];
int pre[msize];
int m,nVertex; //n条边，m个顶点

bool bfs(int s,int t) //寻找一条从s到t的增广路，若找到，返回true
{
    int p;
    queue<int> Q;
    memset(pre,-1,sizeof(pre));
    memset(visited,false,sizeof(visited));

    pre[s]=s;
    visited[s]=true;
    Q.push(s);

    while (!Q.empty())
    {
        p=Q.front(),Q.pop();
        for (int i=1; i<=nVertex; i++)
        {
            if (r[p][i]>0&&!visited[i])
            {
                pre[i]=p;
                visited[i]=true;
                if (i==t) return true;
                Q.push(i);
            }
        }
    }

    return false;
}

int maxFlow(int s,int t)
{
    int flow=0,d;

    while (bfs(s,t))
    {
        d=INF;
        for (int i=t; i!=s; i=pre[i]) d=min(d,r[pre[i]][i]);
        for (int i=t; i!=s; i=pre[i]) r[pre[i]][i] -= d, r[i][pre[i]] += d;
        flow += d;
    }
    return flow;
}

int main()
{
    freopen("telecow.in","r",stdin);
    freopen("telecow.out","w",stdout);
    int s,e,c;

    cin>>nVertex>>m>>s>>e;
    nVertex*=2;
    for(int i=0;i<m;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        r[a*2-1][a*2]=1;
        r[b*2-1][b*2]=1;
        r[a*2][b*2-1]=INF;
        r[b*2][a*2-1]=INF;
    }
    memcpy(origin,r,sizeof r);
    int maxflow=maxFlow(s*2,e*2-1);
    int sum=maxflow;
    memcpy(r,origin,sizeof r);
    printf("%d
",maxflow);

    bool first=true;
    int cnt=0;
    for(int i=1;i<=nVertex/2;i++) // 模拟删掉第i个点
    {
        if(i==s || i==e)
            continue;
        if(cnt==sum)
        {
            break;
        }
        memcpy(origin,r,sizeof r);
        r[i*2-1][i*2]=0;

        if(maxFlow(s*2,e*2-1)+1==maxflow)
        {
            maxflow--;
            cnt++;
            if(first)
            {
                first=false;
            }
            else
            {
                printf(" ");
            }
            printf("%d",i);
            memcpy(r,origin,sizeof r);
            r[i*2-1][i*2]=0;
        }
        else
        {
            memcpy(r,origin,sizeof r);
        }
    }
    cout<<endl;
    return 0;
}