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; }