hdu1083 Courses--最大匹配 & HK算法

来源：互联网发布：最新定位软件编辑：程序博客网时间：2024/06/06 16:59

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1083

题意：p个课程，n个学生，每门课程都有若干个学生，问是否可以让每门课程都有课代表，可以输出YES。

直接用HK算法就可以，当然这题有点水，用匈牙利也可以。

#define _CRT_SECURE_NO_DEPRECATE #include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#define INF 99999999;using namespace std;int t;int dis;int p, n;int a[105][305];int cx[305];int cy[305];int dx[305];int dy[305];bool vis[305];bool searchPath(){queue<int> Q;memset(dx, -1, sizeof(dx));memset(dy, -1, sizeof(dy));dis = INF;for (int i = 1; i <= p; i++){if (cx[i] == -1){Q.push(i);dx[i] = 0;}}while (!Q.empty()){int u = Q.front();Q.pop();if (dx[u] > dis)break;for (int v = 1; v <= n; v++){if (a[u][v] && dy[v] == -1){dy[v] = dx[u] + 1;if (cy[v] == -1)dis = dy[v];else{dx[cy[v]] = dy[v] + 1;Q.push(cy[v]);}}}}return dis != INF;}int findPath(int u){for (int v = 1; v <= n; v++){if (!vis[v] && a[u][v] && dy[v] == dx[u] + 1){vis[v] = 1;if (cy[v] != -1 && dy[v] == dis)continue;if (cy[v] == -1 || findPath(cy[v])){cy[v] = u;cx[u] = v;return 1;}}}return 0;}int maxMatch(){int ans = 0;memset(cx, -1, sizeof(cx));memset(cy, -1, sizeof(cy));while (searchPath()){memset(vis, 0, sizeof(vis));for (int i = 1; i <= p; i++){if (cx[i] == -1)ans += findPath(i);}}return ans;}int main(){scanf("%d", &t);while (t--){scanf("%d%d", &p, &n);memset(a, 0, sizeof(a));for (int i = 1; i <= p; i++){int num, x;scanf("%d", &num);for (int j = 1; j <= num; j++){scanf("%d", &x);a[i][x] = 1;}}printf("%s\n", maxMatch() == p ? "YES" : "NO");}return 0;}

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