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HDU 5090 二分最大匹配

2024-08-04 22:03:13   218人阅读

Game with Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 646    Accepted Submission(s): 284


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）
 
 
题目意思：
给一序列，序列每个数字可以加上0个或多个k，若这个序列能形成每个i有a[i]==i，输出jerry，否则输出tom。
 
思路：
原序列中若a变成c，那么b就不能变成c了，这样就满足二分匹配了。a可以变成a+j*k（j>=0&&j<=n） ，a和a+j*k连边建图求二分最大匹配，若匹配数ans==n输出jerry，否则输出tom。
 
 
代码：
 1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <iostream> 5 #include <vector> 6 #include <queue> 7 using namespace std; 8  9 #define N 10510 11 int max(int x,int y){return x>y?x:y;}12 int min(int x,int y){return x<y?x:y;}13 int abs(int x,int y){return x<0?-x:x;}14 15 int a[N];16 int n, m, k;17 int from[N];18 bool visited[N];19 vector<int>ve[N];20 21 int march(int u){22     int i, v;23     for(i=0;i<ve[u].size();i++){24         v=ve[u][i];25         if(!visited[v]){26             visited[v]=true;27             if(from[v]==-1||march(from[v])){28                 from[v]=u;29                 return 1;30             }31         }32     }33     return 0;34 }35 36 main()37 {38     int i, j;39     cin>>m;40     while(m--){41         scanf("%d %d",&n,&k);42         for(i=1;i<=n;i++) scanf("%d",&a[i]);43         for(i=0;i<=n;i++) ve[i].clear();44         bool flag;45         for(i=1;i<=n;i++){46             flag=false;47             for(j=0;j<=i;j++){48                 if(a[i]+j*k<=n&&(a[i]+j*k)) {49                     ve[i].push_back(a[i]+j*k);50                 }51                 else break;52             }53         }54         55         memset(from,-1,sizeof(from));56         int ans=0;57         for(i=1;i<=n;i++){58             memset(visited,false,sizeof(visited));59             if(march(i)) ans++;60         }61         if(ans!=n) printf("Tom\n");62         else printf("Jerry\n");63     }64 }

 

HDU 5090 二分最大匹配


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