POJ 1469 COURSES（匈牙利算法二分图最大匹配）

COURSES

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 20333		Accepted: 7992

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

Source

大体题意：

给你P个课程，并且给出每个课程的学生，求出学生与课程的最大匹配数目，问结果是否与课程数目相同，相同输出YES，否则NO

思路：

把课程看作左结点，学生看成右结点，就是二分图的最大匹配数目。匈牙利算法即可！

#include
#include
#include
#include
using namespace std;
const int maxn = 300 + 10;
int n;
vectorg[maxn];
int from[maxn], tot;
bool use[maxn];
bool match(int x){
	int len = g[x].size();
	for (int i = 0; i < len; ++i)
	if (!use[g[x][i]]){
		use[g[x][i]] = true;
		if (from[g[x][i]] == -1 || match(from[g[x][i]])){
			from[g[x][i]] = x;
			return true;
		}
	}
	return false;	
}
int hungary(){
	tot = 0;
	memset(from, 255,sizeof from);
	for (int i = 1 ; i <= n; ++i){
		memset(use,0,sizeof use);
		if (match(i))++tot;
	}
	return tot;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		for (int i = 0; i < maxn; ++i)g[i].clear();
		int m,u,v;
		scanf("%d%d",&n,&m);
		for (int i = 1; i <= n; ++i){
			scanf("%d",&u);
			while(u--){
				scanf("%d",&v);
				g[i].push_back(v);
			}
		}
		hungary();
		bool ok = true;
		if (tot != n)ok = false;
		printf("%s",ok?"YES\n":"NO\n");
	}
	return 0;
}