Poj(2771)，最大独立集

最新推荐文章于 2021-05-14 12:29:22 发布

转载最新推荐文章于 2021-05-14 12:29:22 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/TreeDream/p/5761859.html

本文解析了一道经典的二分图匹配问题，通过构建二分图并运用匈牙利算法求解最大匹配数，进而得出最大独立集的规模。题目要求根据学生信息筛选出无法形成情侣的最大人数。

题目链接：http://poj.org/problem?id=2771

Guardian of Decency

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 5517		Accepted: 2322

Description

Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of them might become couples. While you can never exclude this possibility, he has made some rules that he thinks indicates a low probability two persons will become a couple:

Their height differs by more than 40 cm.
They are of the same sex.
Their preferred music style is different.
Their favourite sport is the same (they are likely to be fans of different teams and that would result in fighting).

So, for any two persons that he brings on the excursion, they must satisfy at least one of the requirements above. Help him find the maximum number of persons he can take, given their vital information.

Input

The first line of the input consists of an integer T ≤ 100 giving the number of test cases. The first line of each test case consists of an integer N ≤ 500 giving the number of pupils. Next there will be one line for each pupil consisting of four space-separated data items:

an integer h giving the height in cm;
a character 'F' for female or 'M' for male;
a string describing the preferred music style;
a string with the name of the favourite sport.

No string in the input will contain more than 100 characters, nor will any string contain any whitespace.

Output

For each test case in the input there should be one line with an integer giving the maximum number of eligible pupils.

Sample Input

2
4
35 M classicism programming
0 M baroque skiing
43 M baroque chess
30 F baroque soccer
8
27 M romance programming
194 F baroque programming
67 M baroque ping-pong
51 M classicism programming
80 M classicism Paintball
35 M baroque ping-pong
39 F romance ping-pong
110 M romance Paintball

Sample Output

3
7

Source

Northwestern Europe 2005

看到题意我笑了。

题意：人和人之间如果满足4个条件的任意一条就不会成为夫妻，给定一些人，问从中最多能选多少人且保证任意两人不可能成为夫妻。

分析：由于条件之一是性别相同则不能成为夫妻。我们根据性别把人划分为两个集合，每个人是一个结点构成二分图，若两个人可能成为夫妻则连一条边。然后最大独立集 = 顶点数 - 最大匹配数。

#include <stdio.h>
#include <string.h>
#include <math.h>

bool maps[550][550];
bool use[550];
int match[550];

struct Stu {
    int h;
    char sex[2];
    char music[50];
    char sport[50];
}mans[550],womans[550];

int nman,nwoman;

bool judge(Stu a,Stu b)
{
    if(strcmp(a.music,b.music)) return false;
    if(!strcmp(a.sport,b.sport)) return false;
    if(fabs((a.h-b.h)*1.0)>40) return false;
    return true;
}

bool DFS(int u)
{
    for(int i=0;i<nwoman;i++)
    {
        if(!use[i]&&maps[u][i])
        {
            use[i] = true;
            if(match[i]==-1||DFS(match[i]))
            {
                match[i] = u;
                return true;
            }
        }
    }
    return false;
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(maps,false,sizeof(maps));
        memset(match,-1,sizeof(match));
        int num;
        nman = 0;
        nwoman = 0;
        scanf("%d",&num);
        for(int i=0;i<num;i++)
        {
            int height;
            char sex[2];
            char musics[50];
            char sports[50];
            scanf("%d%s%s%s",&height,sex,musics,sports);
            if(sex[0]=='M')
            {
                mans[nman].h = height;
                strcpy(mans[nman].sex,sex);
                strcpy(mans[nman].music,musics);
                strcpy(mans[nman++].sport,sports);
            }
            if(sex[0]=='F')
            {
                womans[nwoman].h = height;
                strcpy(womans[nwoman].sex,sex);
                strcpy(womans[nwoman].music,musics);
                strcpy(womans[nwoman++].sport,sports);
            }
        }

        for(int i=0;i<nman;i++)
        {
            for(int j=0;j<nwoman;j++)
            {
                if(judge(mans[i],womans[j]))
                    maps[i][j] = true;
            }
        }

        int ans = 0;
        for(int i=0;i<nman;i++)
        {
            memset(use,false,sizeof(use));
            if(DFS(i))
                ans++;
        }
        printf("%d\n",num-ans);

    }
    return 0;
}

转载于:https://www.cnblogs.com/TreeDream/p/5761859.html