1．链接地址
https://vjudge.net/problem/POJ-1469
2．问题描述
Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:
输入样例
2 3 3 3 1 2 3 2 1 2 1 1 3 3 2 1 3 2 1 3 1 1
输出样例
YES NO
3．解题思路
P N(课程数跟学生数)
接着有P行,格式为Count studenti studenti+1 ……studentcount
(Count表示对课程1感兴趣的学生数,接着有Count个学生)
如第一行3 1 2 3表示学生1跟学生2跟学生3对课程1感兴趣
输出为:
若能满足上面两个要求这输出”YES”,否则为”NO”
可以直接使用匈牙利算法
4．算法实现源代码
#include<cstdio> #include<iostream> #include<string.h> #include<cmath> #include<cstdio> #define INF 0x3f3f3f3f #define N 310 int G[N][N], vis[N], used[N]; int n, p; bool Find(int u) { int i; for(i = 1 ; i <= n ; i++) { if(!vis[i] && G[u][i]) { vis[i] = 1; if(!used[i] || Find(used[i])) { used[i] = u; return true; } } } return false; } int main() { int a, b, i, ans, t; scanf("%d", &t); while(t--) { scanf("%d%d", &p, &n); memset(G, 0, sizeof(G)); ans = 0; for(i = 1 ; i <= p ; i++) { scanf("%d", &a); while(a--) { scanf("%d", &b); G[i][b] = 1; } } memset(used, 0, sizeof(used)); for(i = 1 ; i <= p ; i++) { memset(vis, 0, sizeof(vis)); if(Find(i)) ans++; } if(ans == p) printf("YES\n"); else printf("NO\n"); } return 0; }
原文：https://www.cnblogs.com/KasenBob/p/11529692.html