UVA 2039 Pets(网络流)

 

 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;

}