fzu 2039 Pets(网络流)

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本文介绍了一个基于网络流算法解决的宠物买卖匹配问题。问题描述为：有n位顾客和m个宠物，每名顾客只能购买一个自己感兴趣的宠物，且部分顾客对某些宠物不感兴趣。通过构建网络流模型并运用增广路径算法，最终求解出可以成功售出的最大宠物数量。

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Problem Description

Are you interested in pets? There is a very famous pets shop in the center of the ACM city. There are totally m pets in the shop, numbered from 1 to m. One day, there are n customers in the shop, which are numbered from 1 to n. In order to sell pets to as more customers as possible, each customer is just allowed to buy at most one pet. Now, your task is to help the manager to sell as more pets as possible. Every customer would not buy the pets he/she is not interested in it, and every customer would like to buy one pet that he/she is interested in if possible.

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

12 2 21 22 1

Sample Output

Case 1: 2

Source

2011年全国大学生程序设计邀请赛（福州）

题意：n只顾客，m个宠物，e种条件，条件表示顾客x不会买宠物y，每个顾客只买一只宠物。求最多卖出几只宠物

思路：网络流，

代码：

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;

#define INF 0x3f3f3f3f
const int N = 205;
int T, n, m, e, s, t;
int g[N][N], f[N][N], p[N], a[N];

void init() {
    memset(g, 0, sizeof(g));
    memset(f, 0, sizeof(f));
    scanf("%d%d%d", &n, &m, &e);
    s = 0; t = n + m + 1;
    for (int i = 1; i <= n; i ++)
	g[s][i] = 1;
    for (int i = n + 1; i <= n + m; i ++)
	g[i][t] = 1;
    for (int i = 1; i <= n; i ++)
	for (int j = n + 1; j <= n + m; j ++)
	    g[i][j] = 1;
    int u, v;
    for (int i = 0; i < e; i ++) {
	scanf("%d%d", &u, &v);
	g[u][v + n] = 0;
    }
}

int solve() {
    queue<int>q;
    int F = 0;
    while (1) {
	memset(a, 0, sizeof(a));
	a[s] = INF;
	q.push(s);
	while (!q.empty()) {
	    int u = q.front(); q.pop();
	    for (int v = 1; v <= t; v ++) {
		if (!a[v] && g[u][v] - f[u][v] > 0) {
		    a[v] = min(a[u], g[u][v] - f[u][v]);
		    q.push(v); p[v] = u;
		}
	    }
	}
	if (a[t] == 0) break;
	for (int v = t; v; v = p[v]) {
	    f[p[v]][v] += a[t];
	    f[v][p[v]] -= a[t];
	}
	F += a[t];
    }
    return F;
}
int main() {
    int cas = 0;
    scanf("%d", &T);
    while (T--) {
	init();
	printf("Case %d: %d\n", ++cas, solve());
    }
    return 0;
}