HDU1016 Prime Ring Problem （DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46702    Accepted Submission(s): 20639

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

 

思路：对每个数进行访问标记，一道简单的DFS题目、

#include <stdio.h>#include <string.h>int f[45],ans = 1,vis[25];void dfs(int cur, int n, int a[]){if(cur == n && !f[a[cur] + a[1]]){for(int i = 1; i <= n; i ++){if(i > 1)putchar(' ');printf("%d",a[i]);}putchar('\n');}for(int i = 2; i <= n; i ++){if(cur < n && !vis[i]){if(!f[i + a[cur]]){a[cur + 1] = i;vis[i] = 1;dfs(cur+1,n,a);vis[i] = 0;}}}}int main(){int n,a[25];f[0] = f[1] = 1;for(int i = 2; i <= 40; i ++){int flag = 1;for(int j = 2; j <= i / 2; j ++){if(i % j == 0){flag = 0;break;}}if(!flag)f[i]  = 1;}while(~scanf("%d",&n)){memset(a,0,sizeof(a));memset(vis,0,sizeof(vis));vis[1] = 1; a[1] = 1;printf("Case %d:\n",ans++);dfs(1,n,a);putchar('\n');}return 0;}