poj 2289 Jamie's Contact Groups 多重匹配

最新推荐文章于 2020-12-25 04:14:30 发布

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Jamie希望简化查找朋友电话号码的过程，通过将联系人列表分组并最小化最大组的大小来提高效率。遵循建议，Jamie提供了一个包含朋友姓名、所属组和可能的多个组的列表。本文档介绍了如何编写程序实现这一目标。

Description

Jamie is a very popular girl and has quite a lot of friends, so she always keeps a very long contact list in her cell phone. The contact list has become so long that it often takes a long time for her to browse through the whole list to find a friend's number. As Jamie's best friend and a programming genius, you suggest that she group the contact list and minimize the size of the largest group, so that it will be easier for her to search for a friend's number among the groups. Jamie takes your advice and gives you her entire contact list containing her friends' names, the number of groups she wishes to have and what groups every friend could belong to. Your task is to write a program that takes the list and organizes it into groups such that each friend appears in only one of those groups and the size of the largest group is minimized.

Input

There will be at most 20 test cases. Ease case starts with a line containing two integers N and M. where N is the length of the contact list and M is the number of groups. N lines then follow. Each line contains a friend's name and the groups the friend could belong to. You can assume N is no more than 1000 and M is no more than 500. The names will contain alphabet letters only and will be no longer than 15 characters. No two friends have the same name. The group label is an integer between 0 and M - 1. After the last test case, there is a single line `0 0' that terminates the input.

Output

For each test case, output a line containing a single integer, the size of the largest contact group.

Sample Input

3 2
John 0 1
Rose 1
Mary 1
5 4
ACM 1 2 3
ICPC 0 1
Asian 0 2 3
Regional 1 2
ShangHai 0 2
0 0

Sample Output

2
2

#include<cstdio>
#include<cstring>
int n,m;//X和Y
bool mat[1001][501],used[501];//mat初始清零
int link[1001][501];
int mid;
bool can(int t)
{
    for(int i=1;i<=m;i++)
    {
        if(!used[i]&&mat[t][i])
        {
            used[i]=1;
            if(link[i][0]<mid)
            {
                link[i][++link[i][0]]=t;
                return true;
            }
            else
            {
                for(int j=1;j<=link[i][0];j++)
                if(can(link[i][j]))
                {
                    link[i][j]=t;
                    return true;
                }
            }
        }
    }
    return false;
}
bool Match()//X中所有点都匹配
{
    for(int i=0;i<=m;i++) link[i][0]=0;
    for(int i=1;i<=n;i++)
    {
        memset(used,false,sizeof(used));
        if(!can(i)) return false;
    }
    return true;
}
int main()
{
    char str[20];
    int x;
    char c;
    while(scanf("%d%d",&n,&m)==2&&n)
    {
        memset(mat,false,sizeof(mat));
        for(int i=1;i<=n;i++)
        {
            scanf("%s",str);
            while(scanf("%d%c",&x,&c))
            {
                mat[i][x+1]=true;
                if(c=='\n') break;
            }
        }
        int low=1,high=n;
        while(low<high)
        {
            mid=(low+high)>>1;
            if(Match())
            {
                high=mid;
            }
            else low=mid+1;
        }
        printf("%d\n",high);
    }
    return 0;
}