hdu 5090 Game with Pearls（最大匹配）

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 196    Accepted Submission(s): 120

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

 

题意：Jerry 和 Tom 玩一个游戏 ， 给你 n 个盒子 ， a[ i ] 表示开始时 ，第 i 个盒子中的小球的个数 。 然后 Jerry 可以在每个

盒子里加入 0 或 k的倍数的小球 ， 操作完后，Jerry 可以重新排列 盒子的顺序，最终使 第 i 个盒子中有 i 个小球。 若Jerry能

使最终的盒子变成那样，就输出 “Jerry” ，否则 输出 “Tom” 。 

思路： 最大匹配，以样例二为例 ：

然后求出最大匹配数cnt ，若 cnt = = n ，输出 “Jerry” ，否则 输出 “Tom” 。 

#include 
#include 
#include 
using namespace std;
const int maxn = 110 ;

bool con[maxn][maxn],vis[maxn];
int match[maxn],n,k,x;

void initial()
{
    memset(con,0,sizeof(con));
    memset(match,-1,sizeof(match));
}

void input()
{
    scanf("%d %d",&n,&k);
    for(int i=1;i<=n;i++)
    {
         scanf("%d",&x);
         while(x<=n)
         {
              con[x][i]=1;
              x+=k;
         }
    }
}

bool dfs(int x)
{
    for(int i=1;i<=n;i++)
    {
        if(!vis[i] && con[x][i])
        {
            vis[i]=1;
            if(match[i]==-1 || dfs(match[i]))
            {
                 match[i]=x;
                 return true;
            }
        }
    }
    return false;
}

void solve()
{
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
         memset(vis,0,sizeof(vis));
         if(dfs(i))  cnt++;
    }
    if(cnt==n)   printf("Jerry\n");
    else   printf("Tom\n");
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        initial();
        input();
        solve();
    }
    return 0;
}