Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18110 Accepted Submission(s): 8109

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

Input

n (0 < n < 20).

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

Sample Input

6
8

Sample Output

Case 1:
4 3 2 5 6
6 5 2 3 4

Case 2:
2 3 8 5 6 7 4
2 5 8 3 4 7 6
4 7 6 5 8 3 2
6 7 4 3 8 5 2

Source

Asia 1996, Shanghai (Mainland China)

Recommend

JGShining

一开始想到用BFS搜索，但是标志设定是个大问题，想了很久还是没想出解决方法，经同学指点之后，我忽然明白，DFS可以解决这个问题，果断换上DFS，WA了几次，对比发现那case没大写。。。。囧

贴下龌龊的代码：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<math.h>
 4 int vis[25], pr[104];
 5 int n;
 6 void prime()
 7 {
 8     int m = 10, i, j;
 9     memset(pr, 0, sizeof(pr));
10     for(i = 2; i <= m; i++)
11     {
12         if(!pr[i])
13         {
14             for(j = i * i; j <= 100; j += i)
15                 pr[j] = 1;
16         }
17     }
18 }
19 void dfs(int g[], int cur)
20 {
21     int i, j;
22     for(i = 2; i <= n; i++)
23     {
24         if(vis[i] || pr[g[cur - 1] + i])continue;
25         g[cur] = i;
26         if(cur + 1 == n)
27         {
28             if(pr[g[cur] + 1])continue;
29             j = 0;
30             printf("%d", g[j]);
31             for(j = 1; j <= cur; j++)
32                 printf(" %d", g[j]);
33             printf("\n");
34             return ;
35         }
36         vis[i] = 1;
37         dfs(g, cur + 1);
38         vis[i] = 0;
39     }
40 }
41 int main()
42 {
43     int count = 0, g[30];
44     prime();
45     while(scanf("%d", &n) != EOF)
46     {
47         printf("Case %d:\n", ++count);
48         memset(g, 0, sizeof(g));
49         memset(vis, 0, sizeof(vis));
50         g[0] = 1;
51         dfs(g, 1);
52         printf("\n");
53     }
54     return 0;
55 }