HDU 5090 Game with Pearls（二分匹配）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5090

Problem Description

Tom and Jerry are playing a game with tubes and pearls. The rule of the game is:

1) Tom and Jerry come up together with a number K.

2) Tom provides N tubes. Within each tube, there are several pearls. The number of pearls in each tube is at least 1 and at most N.

3) Jerry puts some more pearls into each tube. The number of pearls put into each tube has to be either 0 or a positive multiple of K. After that Jerry organizes these tubes in the order that the first tube has exact one pearl, the 2nd tube has exact 2 pearls,
…, the Nth tube has exact N pearls.

4) If Jerry succeeds, he wins the game, otherwise Tom wins.

Write a program to determine who wins the game according to a given N, K and initial number of pearls in each tube. If Tom wins the game, output “Tom”, otherwise, output “Jerry”.



Input

The first line contains an integer M (M<=500), then M games follow. For each game, the first line contains 2 integers, N and K (1 <= N <= 100, 1 <= K <= N), and the second line contains N integers presenting the number of pearls in each tube.



Output

For each game, output a line containing either “Tom” or “Jerry”.



Sample Input

2 
5 1 
1 2 3 4 5 
6 2 
1 2 3 4 5 5

Sample Output

Jerry 
Tom

Source

2014上海全国邀请赛——题目重现（感谢上海大学提供题目）



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题意：

有 n 个容器，每个里面有一些珍珠。

可以在任意容器中添加 k 的倍数个珍珠。

问最终是否能使得每个容器分别有1 ~ n颗珍珠。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
#define MAXN 177
int N;
int g[MAXN][MAXN], linker[MAXN];
bool used[MAXN];
int dfs(int L)//从左边开始找增广路径
{
    int R;
    for(R = 1 ; R <= N ; R++)//这个顶点编号从0开始，若要从1开始需要修改
    {
        if(g[L][R]!=0 && !used[R])
        {
            //找增广路，反向
            used[R]=true;
            if(linker[R] == -1 || dfs(linker[R]))
            {
                linker[R]=L;
                return 1;
            }
        }
    }
    return 0;//这个不要忘了，经常忘记这句
}
int hungary()
{
    int res = 0 ;
    memset(linker,-1,sizeof(linker));
    for(int L = 1; L <= N; L++)
    {
        memset(used,0,sizeof(used));
        if(dfs(L))
            res++;
    }
    return res;
}
int main()
{
    int t;
    int k, res, tt;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&N,&k);
        memset(g,0,sizeof(g));
        for(int i = 1 ; i <= N ; i++ )
        {
            scanf("%d",&tt);
            while(tt <= N)
            {
                g[tt][i] = 1;
                tt+=k;
            }
        }
        res = hungary();
        if(res == N)
        {
            printf("Jerry\n");
        }
        else
        {
            printf("Tom\n");
        }
    }
    return 0 ;
}