ZOJ—— 1137 Girls and Boys（二分匹配）

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1137

题目：

Girls and Boys

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Time Limit: 10 Seconds       Memory Limit: 32768 KB

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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.

An example is given in Figure 1.

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

Output

5
2 

题目描述：

    最大独立集，先二分匹配（匈牙利算法）求出匹配数，然后再用顶点数减去匹配数/2即可。

代码：

#include<stdio.h>
#include<string.h> 
int e[1010][1010],bo[1010],ma[1010];
int sum;
int a,b,c;
int dfs(int x)
{
	int i;
	for(i = 0;i < a;i ++){
		if(!bo[i] && e[x][i]){
			bo[i] = 1;
			if(!ma[i] || dfs(ma[i])){
				ma[i] = x;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int i,j,w;
	while(scanf("%d",&a) != EOF){
		memset(e,0,sizeof(e));
		for(i = 0;i < a;i ++){
			scanf("%d: (%d)", &w, &b);
			for(j = 0;j < b;j++){
				scanf("%d",&c);
				e[w][c] = 1;
			}
		}
		sum = 0;
		memset(ma,0,sizeof(ma));
		for(j = 0;j < a;j ++){
			memset(bo,0,sizeof(bo));
			if(dfs(j))
				sum ++;
		}
		printf("%d\n",a - sum/2);
	}
	return 0;
}