Courses（hdu 1083)

題目描述：

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:

. 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

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

P N
Count1 Student1 1 Student1 2 ... Student1 Count1
Count2 Student2 1 Student2 2 ... Student2 Count2
......
CountP StudentP 1 StudentP 2 ... StudentP CountP

輸入：

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'll 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.
 

輸出：

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.

樣例輸入：

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

樣例輸出：

YES
NO 

code:

很明顯，這是一個入門級的二分圖匹配問題

需要使用模板即可

不過要注意輸入的序號

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=650;
int n,m;
vector<int> g[maxn];
int match[maxn];
bool used[maxn];
void init(){
	int number=n+m;
	for(int i=0;i<=number;i++){
		g[i].clear();
	}
}
void add_edge(int u,int v){
	g[u].push_back(v);
	g[v].push_back(u);
}
bool dfs(int v){
	used[v]=true;
	for(int i=0;i<g[v].size();i++){
		int u=g[v][i],w=match[u];
		if(w<0||!used[w]&&dfs(w)){
			match[v]=u;
			match[u]=v;
			return true;
		}
	}
	return false;
}
int findd(){
	int res=0;
	memset(match,-1,sizeof(match));
//	init();
	int number=n+m;
	for(int v=1;v<=number;v++){
		if(match[v]<0){
			memset(used,0,sizeof(used));
			if(dfs(v))res++;
		}
	}
	return res;
}
int main()
{
	int ttt;
	scanf("%d",&ttt);
	while(ttt--){
		scanf("%d%d",&n,&m);
		int t;
		init();
		for(int i=1;i<=n;i++){
			scanf("%d",&t);
			while(t--){
				int jiang;
				scanf("%d",&jiang);
				add_edge(jiang+n,i);
			}
		}
		int what=findd();
		if(what>=n)printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}