Girls and Boys poj1466 二分图最大独立集

最新推荐文章于 2020-05-20 00:11:59 发布

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本文介绍了一道经典的算法题——POJ 1466 Girls and Boys，该题目的核心在于求解最大独立集问题。文章详细解析了题目要求，并提供了AC代码实现，利用二分图匹配的原理来高效解决此问题。

1.题目原文

http://poj.org/problem?id=1466

Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 12119		Accepted: 5409

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

Source

Southeastern Europe 2000

2.解题思路

3.AC代码

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<utility>
#include<queue>
using namespace std;

//二分图匹配模板
#define MAX_V 505

int V;//顶点数
vector<int> G[MAX_V];//图的邻接表表示
int match[MAX_V];//所匹配的顶点
bool used[MAX_V];//DFS中用到的访问标记

//向图中增加一条连接u和v的边
void add_edge(int u,int v)
{
    G[u].push_back(v);
    G[v].push_back(u);
}

//通过DFS寻找增广路
bool dfs(int v)
{
    used[v]=true;
    for(int i=0;i<G[v].size();i++){
        int u=G[v][i],w=match[u];
        if(w<0||!used[w]&&dfs(w)){
            match[v]=u;
            match[u]=v;
            return true;
        }
    }
    return false;
}

//求解二分图的最大匹配
int bipartite_matching()
{
    int res=0;
    memset(match,-1,sizeof(match));
    for(int v=0;v<V;v++){
        if(match[v]<0){
            memset(used,0,sizeof(used));
            if(dfs(v)){
                res++;
            }
        }
    }
    return res;
}
//二分图匹配模板
int main()
{
    while(scanf("%d",&V)!=EOF)
    {
        for(int i=0;i<V;i++){
            G[i].clear();
        }
        for (int i = 0; i < V; i++)
		{
			int v, m, u;
			scanf("%d: (%d)", &v, &m);
			//这个感觉好六，看来C语言或许有必要学习一下……
			for (int j = 0; j < m; ++j)
			{
				scanf("%d", &u);
				G[v].push_back(u);
			}
		}
		printf("%d\n", V - bipartite_matching());
    }
    return 0;
}