POJ2914 Minimum Cut（最小割）

最新推荐文章于 2019-07-30 15:26:19 发布

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本文介绍了一种解决图论中最小割问题的方法。通过输入包含多个顶点和边的无向图，算法能够找出使图分裂成两个子图所需的最少边数。文中提供了一个具体的实现案例，展示了如何通过迭代寻找最小割。

Minimum Cut

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 9715		Accepted: 4085
Case Time Limit: 5000MS

Description

Given an undirected graph, in which two vertices can be connected by multiple edges, what is the size of the minimum cut of the graph? i.e. how many edges must be removed at least to disconnect the graph into two subgraphs?

Input

Input contains multiple test cases. Each test case starts with two integers N and M (2 ≤ N ≤ 500, 0 ≤ M ≤ N × (N − 1) ⁄ 2) in one line, where N is the number of vertices. Following are M lines, each line contains M integers A, B and C (0 ≤ A, B < N, A ≠ B, C > 0), meaning that there C edges connecting vertices A and B.

Output

There is only one line for each test case, which contains the size of the minimum cut of the graph. If the graph is disconnected, print 0.

Sample Input

Sample Output

2
1
2

Source

Baidu Star 2006 Semifinal
Wang, Ying (Originator)
Chen, Shixi (Test cases)

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题目的意思是给出一张图，求整体最小割

思路：模板题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 800


int mp[MAXN][MAXN],w[MAXN],vis[MAXN],combine[MAXN];
int minC,cnt,n,m;


void Find(int &s,int &t)
{
    memset(vis,0,sizeof(vis));
    memset(w,0,sizeof(w));
    int tmp=INF;
    int q=n;
    while(q--)
    {
        int mx=-INF;
        for(int i=0; i<n; i++)
        {
            if(!vis[i]&&!combine[i]&&mx<w[i])
            {
                mx=w[i];
                tmp=i;
            }
        }
        if(t==tmp)
        {
            minC=w[t];
            return;
        }
        vis[tmp]=1;
        s=t,t=tmp;
        for(int i=0;i<n;i++)
        {
            if(!vis[i]&&!combine[i])
                w[i]+=mp[t][i];
        }
    }
    minC=w[t];
}

int mincut()
{
    int ans=INF;
    int s,t;
    memset(combine,0,sizeof(combine));
    for(int i=0; i<n-1; i++)
    {
        s=t=-1;
        Find(s,t);
        combine[t]=true;
        ans=min(ans,minC);
        for(int i=0;i<n;i++)
        {
            mp[s][i]+=mp[t][i];
            mp[i][s]+=mp[i][t];
        }
    }
    return ans;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(mp,0,sizeof(mp));
        int u,v,w;
        while(m--)
        {
            scanf("%d %d %d",&u,&v,&w);
            mp[u][v]+=w;
            mp[v][u]+=w;
        }
        printf("%d\n",mincut());
    }
    return 0;
}