hdu 1068 Girls and Boys（二分圖求最大獨立集合）

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

            題目大意：給你每個人互相認識的人，然後問最多能找到多少個人都互不認識。其實就是找：最大獨立集合！

已知：二分圖最大獨立集合 = 節點數 - 最大匹配數

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

代碼：

#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);    //輸入格式注意！
            for(j = 0; j < num; j++)
            {
                scanf("%d", &r);
                map[k][r] = true;   //建表時
            }
        }
        sum = 0;
        for(i = 0; i < n; i++)
        {
            memset(flag, false, sizeof(flag));
            sum += find(i);
        }
        sum /= 2;   //二分圖具有對稱性，最大匹配數 /= 2
        //二分圖最大獨立集合 = 節點數 - 最大匹配數
        printf("%d\n", n - sum);
    }

    return 0;
}