HDU 5090 Game with Pearls（贪心, 二部图最大匹配）

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 414    Accepted Submission(s): 212

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）

 

题目大意：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” 。 

解题思路：

方法一、统计后，从小到大分过去，剩下的放到i+k的位置，这样O（n）的可以得到答案。

方法二、转化乘二部图模型，将 初始状态 和 可到达的球数 作为两个集合，符合倍数条件的连边，若最大匹配数=n，则Jerry赢。

#include <iostream>
#include <cstring>
using namespace std;

const int MAXN = 110;
int nCase, n, k, m, cnt[MAXN];

void init() {
    memset(cnt, 0, sizeof(cnt));
}

void input() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        cnt[x]++;
    }
}

void solve() {
    for (int i = 1; i <= n; i++) {
        if (!cnt[i]) {
            cout << "Tom" << endl;
            return;
        }
        cnt[i+k] += cnt[i] - 1;
    }
    cout << "Jerry" << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin >> nCase;
    while (nCase--) {
        init();
        input();
        solve();
    }
    return 0;
}