HDU 5090 Game with Pearls(贪心, 二部图最大匹配)
2014-11-11 20:50
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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; }
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