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
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 105
10
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 }