2771 Guardian of Decency //MaxMatch

 

 

Guardian of Decency

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3126		Accepted: 1308

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

 

 

         将所有可以发生恋爱关系的男女进行配对,即将男同学和女同学做为两个顶点集合,如果某一对男女同学能发生恋爱关系,就将二者连接,从而建立二分图,那么可以带的出去的人数应该等于这个二分图的最大独立集.
         最大独立集:最大的一个集合,其中的每两点之间都不存在边.
         最大独立集=顶点数-最大匹配数.

 

 

#include<iostream>

#include<string.h>

#include<cstdlib>

#include<cmath>

#include<cstdio>

#include<algorithm>

using namespace std;

int usedif[501];

//usedif[i]记录Y顶点子集中编号为i的顶点是否使用,注意Y子集中的最多顶点数为(7-1)*12+12=84

int link[501];//link[i]记录与Y顶点子集中编号为i的顶点相连的X顶点子集中x的编号

int mat[501][501];//mat[i][j]表示顶点i与j之间是否有边

int bn,gn;//gx为X顶点子集中的顶点数目,gy为Y顶点子集中的顶点数目

struct People                                                          //存储学生信息

{

    int h;

    char sex;

    char mu[105];

    char sp[105];

} boy[505],girl[505];

bool can(int t)                                                     //判断Y集合中的顶点是否能与X集合中的顶点t匹配

{

    int i;

    for(i=0;i<gn;i++)

    {

       if(!usedif[i]&&mat[t][i])

       {

            usedif[i]=true;

            if(link[i]==-1||can(link[i]))

            {

                link[i]=t;

                return true;

            }

       }

    }

    return false;

}

int MaxMatch()                                                      //求解最大匹配数

{

    int i,num=0;

    memset(link,-1,sizeof(link));

    for(i=0;i<bn;i++)

    {

        memset(usedif,0,sizeof(usedif));

        if(can(i))

           num++;

    }

    return num;

}

int main()

{

    int i,j,t,n,ans;

    People p;

    scanf("%d",&t);

    while(t--)

    {

         scanf("%d",&n);

         bn=gn=ans=0;

         for(i=0;i<n;i++)

         {

             scanf("%d %c %s %s",&p.h,&p.sex,p.mu,p.sp);

             if(p.sex=='M')

                  boy[bn++]=p;

             else if(p.sex=='F')

                  girl[gn++]=p;

         }

         memset(mat,0,sizeof(mat));

         for(i=0;i<bn;i++)

           for(j=0;j<gn;j++)

              if(abs(boy[i].h-girl[j].h)<=40&&strcmp(boy[i].mu,girl[j].mu)==0&&strcmp(boy[i].sp,girl[j].sp)!=0)

                 mat[i][j]=true;

         ans=n-MaxMatch();

         printf("%d/n",ans);

    }

    return 0;

}