四川第七届 D Vertex Cover（二分图最小点覆盖，二分匹配模板）

Vertex Cover

frog has a graph with nn vertices v(1),v(2),…,v(n)v(1),v(2),…,v(n) and mm edges (v(a1),v(b1)),(v(a2),v(b2)),…,(v(am),v(bm))(v(a1),v(b1)),(v(a2),v(b2)),…,(v(am),v(bm)).

She would like to color some vertices so that each edge has at least one colored vertex.

Find the minimum number of colored vertices.

Input

The input consists of multiple tests. For each test:

The first line contains 22 integers n,mn,m (2≤n≤500,1≤m≤n(n−1)22≤n≤500,1≤m≤n(n−1)2). Each of the following mm lines contains 22 integers ai,biai,bi (1≤ai,bi≤n,ai≠bi,min{ai,bi}≤301≤ai,bi≤n,ai≠bi,min{ai,bi}≤30)

Output

For each test, write 11 integer which denotes the minimum number of colored vertices.

Sample Input

    3 2
    1 2
    1 3
    6 5
    1 2
    1 3
    1 4
    2 5
    2 6

Sample Output

    1
    2

题意：有n个点m条边，每条边至少有一个顶点染色，至少要染多少个点

思路：二分图最小点覆盖，有一个公式 二分图最大匹配=最小点覆盖，匈牙利算法一代就出来了

#include<cstdio>
#include<string.h>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<deque>
using namespace std;
#define ll unsigned long long
vector<int>v[505];
int match[505];
bool used[505];
bool dfs(int x)
{
    used[x]=1;//标记询问过了
    for(int i=0;i<v[x].size();i++)
    {
        int y=v[x][i];
        int w=match[y];
        if(w<0||!used[w]&&dfs(w))
        {//如果它没有匹配过或者（没有问过而且可以找到其它）
            match[x]=y;
            match[y]=x;
            return 1;
        }
    }
    return 0;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=0;i<=502;i++) v[i].clear();
        for(int i=1;i<=m;i++)
        {
            int x,y;
            cin>>x>>y;
            v[x].push_back(y);
            v[y].push_back(x);
        }
        int res=0;
        memset(match,-1,sizeof(match));
        for(int i=1;i<=n;i++)
        {
            if(match[i]<0)//还没匹配
            {
                memset(used,0,sizeof(used));//每次都要初始化
                if(dfs(i))
                //如果能匹配
                    res++;
                }
            }
        }
        cout<<res<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/caiyishuai/p/8955261.html