POJ 1466 Girls and Boys(最大独立集)

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

题意：有n个学生，问有几个学生一点关系也没有，即两个集合中有多少点它们之间一条连线都没有（最大独立集）

   最小顶点覆盖 ： 最大匹配数（在二分图中寻找一个尽量小的点集，使图中每一条边至少有一个点在该点集中）

   反证法证明：假设当前存在一条两个端点都不在最小顶点覆盖点集中，那么这条边肯定可以增大最大匹配的边集，与最大匹配矛盾

   最小路径覆盖：顶点数-最大匹配数（在二分图中寻找一个尽量小的边集，使图中每一个点都是该边集中某条边的端点）一条边最多可以包含两个顶点，所以我们选边的时候让这样的边尽量多，也就是说最大匹配的边集数目咯。剩下的点就只能一个边连上一个点到集合里啦。

最大独立集：顶点数—最大匹配数（除了最大匹配的两个端点，剩下的点肯定是独立的，然后把每个最大匹配的端点取出一个加入独立集，就成了最大独立集）     顶点数-（最大匹配数*2）+最大匹配数 = 顶点数-最大匹配数

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 510
using namespace std;
int G[N][N], n;
int use[N], vis[N];
int Find(int u) //匈牙利算法
{
    int i;
    for (i = 0; i < n; i++)
    {
        if (!vis[i] && G[u][i])
        {
            vis[i] = 1;
            if (!use[i] || Find(use[i]))
            {
                use[i] = u;
                return 1;
            }
        }
    }
    return 0;
}
int main ()
{
    int a, b, m, i, ans, num;
    while (scanf("%d", &n) != EOF)
    {
        memset(G, 0, sizeof(G));
        memset(use, 0, sizeof(use));
        ans = 0;
        for (i = 0; i < n; i++)
        {
            scanf("%d: (%d)", &a, &m);
            while (m--)
            {
                scanf("%d", &b);
                G[a][b] = 1;
            }
        }
        for (i = 0; i < n; i++)
        {
            memset(vis, 0, sizeof(vis));
            if (Find(i))
                ans++;
        }
        num = n-ans/2; //由于这n个学生中有男生也有女生，那么n就是顶点数，2-3和3-2的关系是一样的，所以最大匹配数需要除以2
        printf("%d
", num);
    }
    return 0;
}