hdu 1068 Girls and Boys（二分图求最大独立集合）-优快云博客

本文介绍了一个经典的图论问题——寻找图中的最大独立集合，并通过一个具体的编程实例详细展示了如何使用匈牙利算法来解决该问题。该算法可以有效找出二分图中最大的独立集合，即在图中找到最多数量的顶点，这些顶点之间不存在边相连。

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Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3260 Accepted Submission(s): 1405

Problem Description

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

题目大意：给你每个人互相认识的人，然后问最多能找到多少个人都互不认识。其实就是找：最大独立集合！

已知：二分图最大独立集合 = 节点数 - 最大匹配数

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1068

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

const int N = 1005;

int pre[N];
bool map[N][N], flag[N];
int n;

int find(int cur)   //匈牙利算法
{
    int i;
    for(i = 0; i < n; i++)
    {
        if(map[cur][i] && !flag[i])
        {
            flag[i] = true;
            if(pre[i] == -1 || find(pre[i]))
            {
                pre[i] = cur;
                return 1;
            }
        }
    }
    return 0;
}

int main()
{
    int i, j, r, k, num, sum;
    while(scanf("%d", &n) != EOF)
    {
        memset(map, false, sizeof(map));
        memset(pre, -1, sizeof(pre));
        for(i = 0; i < n ; i++)
        {
            scanf("%d: (%d)", &k, &num);    //输入格式注意！
            for(j = 0; j < num; j++)
            {
                scanf("%d", &r);
                map[k][r] = true;   //建表时
            }
        }
        sum = 0;
        for(i = 0; i < n; i++)
        {
            memset(flag, false, sizeof(flag));
            sum += find(i);
        }
        sum /= 2;   //二分图具有对称性，最大匹配数 /= 2
        //二分图最大独立集合 = 节点数 - 最大匹配数
        printf("%d\n", n - sum);
    }

    return 0;
}