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POJ2771_Guardian of Decency(二分图/最大独立集=N-最大匹配)

2024-07-14 05:51:58   224人阅读

解题报告

http://blog.csdn.net/juncoder/article/details/38159017

题目传送门

题意：

看到题目我就笑了，，，

老师认为这样的两个学生不是一对：

身高相差40以上（年龄都不是距离了，身高又算什么）

不同性别（sad，，，就不允许基友存在呀，，，谁的肥皂掉了，，，）

喜欢不一样的歌曲类型（你总不能要求两人整天听小苹果吧,,,,,,你是我的小丫小苹果,,,,,,）

喜欢一样的运动（ they are likely to be fans of different teams and that would result in fighting......这是什么理由，喜欢同个不同球队还会打起来）

问最多的不在一起的有多少人

思路：

不在一起代表独立于相爱关系的，也就是求最大独立集，集合里面任意两个人都不相爱

最大独立集=结点数-最大匹配

ps不知道是不是写挫了，重写才过，sad。。。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
    int h;
    char sex;
    char mus[110];
    char spo[110];
}M[510],F[510];
int n,m,f,mmap[550][550],pre[550],vis[550];
int pd(int i,int j)
{
    if( abs(M[i].h - F[j].h) > 40 )
        return 0;
    if( strcmp(M[i].mus,F[j].mus) )
        return 0;
    if( strcmp(M[i].spo,F[j].spo) == 0 )
        return 0;
    return 1;
}
int dfs(int x)
{
    for(int i=m+1;i<=m+f;i++){
        if(!vis[i]&&mmap[x][i]){
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i])){
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,t,h;
    char str[100];
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        m=f=1;
        memset(mmap,0,sizeof(mmap));
        memset(pre,-1,sizeof(pre));
        memset(M,0,sizeof(M));
        memset(F,0,sizeof(F));
        for(i=1;i<=n;i++){
            scanf("%d%s",&h,str);
            if(str[0]=='M'){
                scanf("%s%s",M[m].mus,M[m].spo);
                M[m].h=h;
                M[m++].sex='M';
            }
            else {
                scanf("%s%s",F[f].mus,F[f].spo);
                F[f].h=h;
                F[f++].sex='F';
            }
        }
        for(i=1;i<=m;i++){
            for(j=1;j<=f;j++){
                if(pd(i,j))
                mmap[i][m+j]=1;
            }
        }
        int ans=0;
        for(i=1;i<=m;i++){
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",n-ans);
    }
}

Guardian of Decency
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 4876 Accepted: 2056

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


POJ2771_Guardian of Decency(二分图/最大独立集=N-最大匹配)


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