BOJ 5291 Girls and Boys //二分图水题

Girls and Boys

Time Limit: 10000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID:  1068
64-bit integer IO format:  %I64d      Java class name:  Main

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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

Source

Southeastern Europe 2000

#include <stdio.h>
#include <string.h>

#define p 5000

bool tu[p][p];
bool visit[p];
int r[p];

bool dfs(int a, int n)
{
	int i;
	for(i=0; i<n; i++)
	{
		if(tu[a][i] && !visit[i])
		{
			visit[i] = true;
			if(r[i]==-1 || dfs(r[i], n))
			{
				r[i] = a;
				return true;
			}
		}
	}
	return false;
}

int main()
{
	int n;
	int i, a, k, v;
	int ans;

	while(scanf("%d", &n)!=-1)
	{
		ans = 0;
		memset(tu, false, sizeof(tu));
		for(i=0; i<n; i++)
		{
			r[i] = -1;
		}

		for(i=0; i<n; i++)
		{
			scanf("%d: ", &a);
			scanf("(%d)", &k);

			while(k--)
			{
				scanf("%d", &v);
				tu[a][v] = true;
			}
		}

		
		for(i=0; i<n; i++)
		{
			memset(visit, false, sizeof(visit));
			if( dfs(i, n) )
				ans++;
		}
		printf("%d\n", n-ans/2);
	}

	return 0;
}