Girls and Boys

Problem Description

the second year of the university somebody started a study on the romantic relations between the students. The relation “romantically involved” is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been “romantically involved”. The result of the program is the number of students in such a set.

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)

The student_identifier is an integer number between 0 and n-1, for n subjects.
For each given data set, the program should write to standard output a line containing the result.

 

Sample Input

7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0

 

Sample Output

5
2

题解：最大独立集就是在图G中找m个两两不相连的点，最大匹配找完之后，剩下的不就是两两不想连的吗。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

vector<int> vec[1003];
bool visited[1003];
int pre[1003];

bool find(int x)
{
	for(int i = 0; i < vec[x].size();i++)
	{
		int t= vec[x][i];
		if(!visited[t])
		{
			visited[t] = true;
			if(pre[t] == -1 || find(pre[t]))
			{
				pre[t] = x;
				return true;
			}
		}
	}
	
	return false;
}

int main()
{
	int n;
	while(scanf("%d",&n) != EOF)
	{
		int x,y,m;
		memset(pre,-1,sizeof(pre));
		for(int i = 0;i < n;i++)
		{
			vec[i].clear();
		}
		for(int i = 0;i < n;i++)
		{
			scanf("%d: (%d)",&x,&m);
			for(int j = 0;j < m;j++)
			{
				scanf("%d",&y);
				vec[x].push_back(y);
			}
		}
		
		int ans = 0;
		for(int i = 0;i < n;i++)
		{
			memset(visited,false,sizeof(visited));
			if(find(i))
			{
				ans++;
			}
		}
		
		printf("%d\n",n - ans / 2);
	}
	
	
	return 0;
}