HDU 3081 Marriage Match II 二分图最大匹配

Marriage Match II

Problem Description

Presumably, you all have known the question of stable marriage match. A girl will choose a boy; it is similar as the game of playing house we used to play when we are kids. What a happy time as so many friends playing together. And it is normal that a fight or a quarrel breaks out, but we will still play together after that, because we are kids.
Now, there are 2n kids, n boys numbered from 1 to n, and n girls numbered from 1 to n. you know, ladies first. So, every girl can choose a boy first, with whom she has not quarreled, to make up a family. Besides, the girl X can also choose boy Z to be her boyfriend when her friend, girl Y has not quarreled with him. Furthermore, the friendship is mutual, which means a and c are friends provided that a and b are friends and b and c are friend.
Once every girl finds their boyfriends they will start a new round of this game—marriage match. At the end of each round, every girl will start to find a new boyfriend, who she has not chosen before. So the game goes on and on.
Now, here is the question for you, how many rounds can these 2n kids totally play this game?

 

Input

There are several test cases. First is a integer T, means the number of test cases.
Each test case starts with three integer n, m and f in a line (3<=n<=100,0<m<n*n,0<=f<n). n means there are 2*n children, n girls(number from 1 to n) and n boys(number from 1 to n).
Then m lines follow. Each line contains two numbers a and b, means girl a and boy b had never quarreled with each other.
Then f lines follow. Each line contains two numbers c and d, means girl c and girl d are good friends.

 

Output

For each case, output a number in one line. The maximal number of Marriage Match the children can play.

 

Sample Input

1

4 5 2

1 1

2 3

3 2

4 2

4 4

1 4

2 3

 

Sample Output

2

 

 

 

过家家游戏，有N个男孩和N个女孩，女孩优先。

女孩可以选择没有和她发生争吵的男孩组成一对，也可以和她的闺蜜中没有和她闺蜜发生争吵的男孩组成一对。

组成N对后就开始下一轮，但每一轮都不可以和前面已经组成的人组成一对，问最多可以玩多少轮。

 

二分最大匹配的模板题，不过边就要用并查集处理下。每次能匹配的人数等于n，ans就加1，然后把已经匹配的边删掉，小与n的话就说明不能匹配了。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 110;
int edge[N][N], match[N], vis[N], fa[N];
int n, m, f, t;
int find(int x) {
    return fa[x] = (fa[x]==x ? x: find(fa[x]));
}
bool dfs(int u) {
    for(int v = 1; v <= n; v ++) {
        if(edge[u][v] && !vis[v]) {
            vis[v] = 1;
            if(match[v] == -1 || dfs(match[v])){
                match[v] = u;
                return true;
            }
        }
    }
    return false;
}
int main() {
    ios::sync_with_stdio(false);
    cin>>t;
    while(t--) {
        cin>>n>>m>>f;
        for(int i = 1; i <= n; i ++) fa[i] = i;
        memset(edge, 0, sizeof(edge));
        int u, v;
        for(int i = 1; i <= m; i ++) {
            cin>>u>>v;
            edge[u][v] = 1;
        }
        for(int i = 1; i <= f; i ++) {
            cin>>u>>v;
            fa[find(u)] = find(v);
        }
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= n; j ++) {
                if(find(i)==find(j)) {
                    for(int k = 1; k <= n; k ++) {
                        if(edge[i][k])edge[j][k] = 1;
                    }
                }
            }
        }
        int ans = 0;
        while(1) {
            int res = 0;
            memset(match, -1, sizeof(match));
            for(int i = 1; i <= n; i ++) {
                memset(vis, 0, sizeof(vis));
                if(dfs(i)) res++;
            }
            if(res < n) break;
            ans++;
            for(int i = 1; i <= n; i ++) {
                if(match[i] != -1) {
                    edge[match[i]][i] = 0;
                }
            }
            // for(int i = 1; i <= n; i ++) {
            //     for(int j = 1; j <= n; j ++) printf("%d ", edge[i][j]);
            //     printf("
");
            // }
        }
        printf("%d
",ans);
    }
    return 0;
}