Girls and Boys
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
题意概括：
算出有几个人没有配对。
解题分析：
这道题是二分匹配中算最大独立集，最大独立集合 = 节点数 - 最大匹配数。
AC代码：
#include<stdio.h>
#include<string.h>
#define N 1005
int n;
int match[N], book[N];
int e[N][N];
int dfs(int u)
{
for(int i = 0; i < n; i++){
if(e[u][i] && !book[i]){
book[i] = 1;
if(match[i] == -1 || dfs(match[i])){
match[i] = u;
return 1;
}
}
}
return 0;
}
int main()
{
int i, u, v, s, sum;
while(~scanf("%d", &n)){
memset(e, 0, sizeof(e));
memset(match, -1, sizeof(match));
for(i = 0; i < n; i++){
scanf("%d: (%d)", &u, &s);
while(s--){
scanf("%d", &v);
e[u][v] = 1;
}
}
for(i = sum = 0; i < n; i++){
memset(book, 0, sizeof(book));
if(dfs(i)) sum++;
}
printf("%d\n", n-sum/2);
}
return 0;
}