HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39699    Accepted Submission(s): 17495

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:

1 4 3 2 5 6

1 6 5 2 3 4

 

Case 2:

1 2 3 8 5 6 7 4

1 2 5 8 3 4 7 6

1 4 7 6 5 8 3 2

1 6 7 4 3 8 5 2

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

 

 

解析：DFS。

 

 

 

#include <cstdio>
#include <cstring>

int n;
bool pri[40] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};
int a[25];
bool vis[25];

void dfs(int level)
{
    if(level == n && pri[a[1]+a[n]]){
        for(int i = 1; i<n; ++i){
            printf("%d ",a[i]);
        }
        printf("%d\n",a[n]);
        return;
    }
    for(int i = 2; i <= n; ++i){
        if(!vis[i] && pri[i+a[level]]){
            a[++level] = i;
            vis[i] = true;
            dfs(level);
            vis[i] = false;
            --level;
        }
    }
}

int main()
{
    memset(vis,0,sizeof(vis));
    int cnt = 0;
    while(~scanf("%d",&n)){
        printf("Case %d:\n",++cnt);
        a[1] = 1;
        dfs(1);
        printf("\n");
    }
    return 0;
}