fzu 2039 Pets（网络流最大流）

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题目大意：fzu 2039 Pets

题目大意：给出n和m，代表有n个人和m只狗，然后给出t表示t个关系，每个关系给出a b，表示第a个人不会买第b条狗。问说商店最多卖出多少条狗。

解题思路：网络流的最大流问题，增广路算法，不同的是建图的过程，每新增一条关系就会删除一条边儿不是增加一条边。

#include <stdio.h>#include <string.h>#include <queue>using namespace std;const int N  = 205;const int INF = 0x3f3f3f3f;int n, m, tmp, t, g[N][N], f[N][N];void init() {memset(g, 0, sizeof(g));memset(f, 0, sizeof(f));scanf("%d%d%d", &n, &m, &t);for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++)g[i][j + n] = 1;}int a, b;for (int i = 0; i < t; i++) {scanf("%d%d", &a, &b);g[a][b + n] = 0;}tmp = n + m + 1;for (int i = 1; i <= n; i++) g[0][i] = 1;for (int i = 1; i <= m; i++) g[n + i][tmp] = 1;}int solve() {queue<int> q;int d[N], v[N], ans = 0;while (true) {memset(d, 0, sizeof(d));memset(v, 0, sizeof(v));d[0] = INF;q.push(0);while ( !q.empty() ) {int u = q.front(); q.pop();for (int i = 0; i <= tmp; i++) {if (g[u][i] - f[u][i] > 0 && !d[i]) {d[i] = min(d[u], g[u][i] - f[u][i]);v[i] = u;q.push(i);}}}if (d[tmp] == 0) break;ans += d[tmp];for (int i = tmp; i; i = v[i]) {f[v[i]][i] += d[tmp];f[i][v[i]] -= d[tmp];}}return ans;}int main () {int cas;scanf("%d", &cas);for (int i = 1; i <= cas; i++) {init();printf("Case %d: %d\n", i, solve() );}return 0;}

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