【二分图|最大匹配】POJ-1469 COURSES

COURSES

Time Limit: 1000MS		Memory Limit: 10000K

Description

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: 

• 
  
• every student in the committee represents a different course (a student can represent a course if he/she visits that course) 
  
• each course has a representative in the committee 
  

Input

Your program should read sets of data from the std input. The first line of the input contains the number of the data sets. Each data set is presented in the following format: 

P N 
Count1 Student _{1 1} Student _{1 2} ... Student _{1 Count1} 
Count2 Student _{2 1} Student _{2 2} ... Student _{2 Count2} 
... 
CountP Student _{P 1} Student _{P 2} ... Student _{P CountP} 

The first line in each data set contains two positive integers separated by one blank: P (1 <= P <= 100) - the number of courses and N (1 <= N <= 300) - the number of students. The next P lines describe in sequence of the courses �from course 1 to course P, each line describing a course. The description of course i is a line that starts with an integer Count i (0 <= Count i <= N) representing the number of students visiting course i. Next, after a blank, you抣l find the Count i students, visiting the course, each two consecutive separated by one blank. Students are numbered with the positive integers from 1 to N. 
There are no blank lines between consecutive sets of data. Input data are correct. 

Output

The result of the program is on the standard output. For each input data set the program prints on a single line "YES" if it is possible to form a committee and "NO" otherwise. There should not be any leading blanks at the start of the line.

Sample Input

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

Sample Output

YES
NO

————————————————————疲れた分割線————————————————————

思路：水题没什么好说的。。要多多练习盲打和Vim！

P.S. 顺便说一句，在这样的二部图中，因为是单向边，所以就算X部和Y部共用点都是没问题的，可以节省空间。

代码如下：

/*
ID: j.sure.1
PROG:
LANG: C++
*/
/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <climits>
#include <iostream>
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
/****************************************/
const int N = 444, M = 33333;
struct Edge {
	int v, next;
	Edge(){}
	Edge(int _v, int _next):
		v(_v), next(_next){}
}edge[M];
int tot, head[N], mac[N], n, m;
bool vis[N];

void init()
{
	tot = 0;
	memset(head, -1, sizeof(head));
}

void add(int u, int v)
{
	edge[tot] = Edge(v, head[u]); head[u] = tot++;
}

bool dfs(int u)
{
	for(int i = head[u]; ~i; i = edge[i].next) {
		int v = edge[i].v;
		if(!vis[v]) {
			vis[v] = true;
			if(mac[v] == -1 || dfs(mac[v]))
			{
				mac[v] = u;
				return true;
			}
		}
	}
	return false;
}

int match()
{
	int ret = 0;
	memset(mac, -1, sizeof(mac));
	for(int i = 1; i <= m; i++) {
		memset(vis, 0, sizeof(vis));
		if(dfs(i)) {
			ret++;
		}
	}
	return ret;
}

int main()
{
#ifdef J_Sure
//	freopen("000.in", "r", stdin);
//	freopen(".out", "w", stdout);
#endif
	int Cas;
	scanf("%d", &Cas);
	while(Cas--) {
		init();
		scanf("%d%d", &m, &n);
		int p;
		for(int i = 1; i <= m; i++) {
			scanf("%d", &p);
			int x;
			while(p--) {
				scanf("%d", &x);
				add(i, x);//X部、Y部共用点
			}
		}
		int ans = match();
		printf("%s\n", ans < m ? "NO" : "YES");
	}
	return 0;
}