hdu1083:Courses(二分图最大匹配)

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

Problem 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

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

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.

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.

An example of program input and output:

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

题意分析:

有n门课程和m个同学,每个课程有多个同学学习,问能不能给每一个课程找一个课代表,每个人只能担任一个课程的课代表。

解题思路:

匈牙利匹配

#include <stdio.h>
#include <vector>
#include <string.h>
#define N 320
using namespace std;
int l[N], book[N], n, m;
vector<int>a[N];
int dfs(int u)
{
	int i, v, s;
	v=a[u].size();
	for(i=0; i<v; i++)
	{
		s=a[u][i];
		if(book[s]==0)
		{
			book[s]=1;
			if(l[s]==0 || dfs(l[s]))
			{
				l[s]=u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int T, i, t, v, sum;
	scanf("%d", &T);
	while(T--)
	{
		memset(l, 0, sizeof(l));
		scanf("%d%d", &n, &m);
		for(i=1; i<=n; i++)
			a[i].clear();
		for(i=1; i<=n; i++)
		{
			scanf("%d", &t);
			while(t--)
			{
				scanf("%d", &v);
				a[i].push_back(v);
			}
		}
		sum=0;
		for(i=1; i<=n; i++)
		{
			memset(book, 0, sizeof(book));
			if(dfs(i))
				sum++;
		}
		if(sum==n)
			printf("YES
");
		else
			printf("NO
");
	}
	return 0;
}
原文地址:https://www.cnblogs.com/zyq1758043090/p/11852648.html