POJ 1466：Girls and Boys 二分图的最大点独立集

Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 11097		Accepted: 4960

Description

In 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.
Input

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 (n <=500 ), for n subjects.
Output

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

有一堆男生和女生，有的男生对女生喜欢，对应的女生也喜欢相应的男生。现在要找到一个集合，里面的人互相都没有感觉，问这个集合中的最大人数。

误打误撞了最大独立集，这道题的二分图的最大独立集=所有节点的数量-最大匹配数量/2。上面的这个也是有要求的，就是喜欢的女生恰好也喜欢你，每两个点上都有两条边，每一次都重复算了一次。要是不是这样，每次喜欢的人的关系不确定喜不喜欢你，我觉得解起来还会麻烦一些。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int grid[805][805];
int link[805];
int visit[805];
int n,k,V1,V2;
int result;

bool dfs(int x)
{
int i;
for(i=0;i<V2;i++)
{
if(grid[x][i]==1&&visit[i]==0)
{
visit[i]=1;
if(link[i]==-1||dfs(link[i]))
{
link[i]=x;
return true;
}
}
}
return false;
}

void Magyarors()
{
int i;

result=0;
memset(link,-1,sizeof(link));//！！这里不能是0

for(i=0;i<V1;i++)
{
memset(visit,0,sizeof(visit));
if(dfs(i))
result++;
}
cout<<n-result/2<<endl;
}

int main()
{
int i,j,temp1,temp2,temp3;
while(scanf("%d",&n)!=EOF)
{
memset(grid,0,sizeof(grid));
V1=V2=n;
for(i=0;i<n;i++)
{
scanf("%d: (%d)",&temp1,&temp2);
for(j=1;j<=temp2;j++)
{
scanf("%d",&temp3);
grid[temp1][temp3]=1;
}
}
Magyarors();
}
return 0;
}