无向图的最大独立子集：Girls ans Boys

最新推荐文章于 2021-02-13 20:03:46 发布

lethic

最新推荐文章于 2021-02-13 20:03:46 发布

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分类专栏： ACM_网络流文章标签： linker output each integer input 64bit

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本文探讨了如何通过编程解决大学校园中学生间浪漫关系的研究问题。利用匈牙利算法求解二分图的最大匹配问题，进而计算出不包含任何曾有过浪漫关系的学生集合的最大数量。

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Girls and Boys Time Limit:5000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u

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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

无向图最小覆盖路径 = 顶点数 - 最大匹配数/2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN=1010;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的
int linker[MAXN];//是否有父亲，如果有，父亲是谁
bool used[MAXN]; //在本次dfs中是否在路径中被指过
bool dfs(int u){
    int v;
    for(v=0;v<vN;v++)
        if(g[u][v]&&!used[v]){
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v])){//v没有父亲或v的父亲有其他儿子
                linker[v]=u;
                return true;
            }
        }
    return false;
}

int hungary(){
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=0;u<uN;u++){
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}

int main(){
    int n;
    int i, j;
    while(~scanf("%d", &n)){
        memset(g, 0, sizeof(g));
        uN = n;
        vN = n;
        for(i = 0; i < n; i ++){
            int a, b;
            int num;
            scanf("%d", &a);
            getchar();
            getchar();
            getchar();
            scanf("%d", &num);
            getchar();
            getchar();
            for(j = 0; j < num; j ++){
                scanf("%d", &b);
                g[a][b] = 1;
                g[b][a] = 1;
             }
        }
        int ans = n - hungary()/2;
        printf("%d\n", ans);
    }
    return 0;
}