hdu3549 Flow Problem--EK算法 & 最大流

来源：互联网发布：win10网络红叉编辑：程序博客网时间：2024/05/01 01:13

原题链接： http://acm.hdu.edu.cn/showproblem.php?pid=3549

题意：给定n个点，m条边，m行数据u,v,w代表边u到v的权值是w，注意是有向边。求最大流。

模板题。

#define _CRT_SECURE_NO_DEPRECATE #include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>using namespace std;int pre[20];int vis[20];int p[20][20];int s, t;int n, m;bool bfs(){memset(pre, 0, sizeof(pre));memset(vis, 0, sizeof(vis));vis[s] = 1;queue<int> Q;Q.push(s);while (!Q.empty()){int cur = Q.front();Q.pop();if (cur == t)return true;for (int i = 1; i <= n; i++){if (!vis[i] && p[cur][i]){Q.push(i);pre[i] = cur;vis[i] = 1;}}}return false;}int maxFlow(){int ans = 0;while (1){if (!bfs())//如果找不到增广路径就返回return ans;int minn = 99999999;for (int i = t; i != s; i = pre[i]) //通过pre[]数组查找增广路径上的边，求出残留容量的最小值minn = min(minn, p[pre[i]][i]);for (int i = t; i != s; i = pre[i]){p[pre[i]][i] -= minn;p[i][pre[i]] += minn;}ans += minn;}}int main(){int T;int u, v, w;scanf("%d", &T);for (int cas = 1; cas <= T; cas++){memset(p, 0, sizeof(p));scanf("%d%d", &n, &m);s = 1;t = n;for (int i = 1; i <= m; i++){scanf("%d%d%d", &u, &v, &w);p[u][v] += w;//注意这里是+号}int ans = maxFlow();printf("Case %d: %d\n", cas, ans);}return 0;}

1 0