题干：

有n个人，m条狗，然后会给出有一些人不喜欢一些狗就不会购买，问最多能卖多少狗。。

Input

There is a single integer T in the first line of the test data indicating that there are T(T≤100) test cases. In the first line of each test case, there are three numbers n, m(0≤n,m≤100) and e(0≤e≤n*m). Here, n and m represent the number of customers and the number of pets respectively.

In the following e lines of each test case, there are two integers x(1≤x≤n), y(1≤y≤m) indicating that customer x is not interested in pet y, such that x would not buy y.

Output

For each test case, print a line containing the test case number (beginning with 1) and the maximum number of pets that can be sold out.

Sample Input

1
2 2 2
1 2
2 1

Sample Output

Case 1: 2

解题报告：

   随便匈牙利一下。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e2 + 5;
int nxt[MAX];
bool line[MAX][MAX];
bool use[MAX];
int n,m,e;
bool find(int x) {
	for(int i = 1; i<=m; i++) {
		if(!use[i]&& line[x][i]) {
			use[i]=1;
			if(!nxt[i] || find(nxt[i])) {
				nxt[i] = x;
				return 1;
			}
		}
	}
	return 0;
}

int match() {
	int res = 0 ;
	for(int i = 1; i<=n; i++) {
		memset(use,0,sizeof use);
		if(find(i)) res++;
	}
	return res;
}
int main()
{
	int t,iCase=0;
	cin>>t;
	while(t--) {
		memset(line,1,sizeof line);
		memset(nxt,0,sizeof nxt);
		scanf("%d%d%d",&n,&m,&e);
		for(int a,b,i = 1; i<=e; i++) {
			scanf("%d%d",&a,&b);
			line[a][b]=0;
		}
		printf("Case %d: %d\n",++iCase,match());
	}


	return 0 ;
}