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 <iostream>#include <cstdio>#include <cstring>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;}