HDOJ 1016 Prime Ring Problem （DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36982    Accepted Submission(s): 16305

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

 

#include#includeint prime[40] = {0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};   //素数表int visit[21];    //访问标记int a[21];    // 存放序列结果int n;void dfs(int num){    int i;    if (num == n && prime[a[num - 1] + 1]){   //结束条件        for (i = 0; i < num - 1; i++)            printf("%d ", a[i]);        printf("%d\n", a[num - 1]);    }    else{        for (i = 2; i <= n; i++){            if (!visit[i] && prime[i + a[num - 1]]){   //如果i没有被用过并且能与前面的数和成素数                visit[i] = 1;  //标记i已被访问                a[num++] = i;  //将i加入结果序列                dfs(num);      //递归访问                visit[i] = 0;  //回溯法标准框架去掉标记                num--;            }        }    }}int main(){    int num = 0;    while(scanf("%d", &n) != EOF){        num++;        printf("Case %d:\n", num);        memset(visit, 0, sizeof(visit));        a[0] = 1;        dfs(1);        printf("\n");    }    return 0;}