hdu 1016 Prime Ring Problem

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：使用深搜遍历。

 

代码：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int prim[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 n,vis[21];void dfs(int num,int dij[]){    if(num>n&&prim[dij[n]+dij[1]])    {        printf("%d",dij[1]);        for(int i=2; i<=n; i++)            printf(" %d",dij[i]);        printf("\n");    }    else        for(int i=2; i<=n; i++)            if(!vis[i]&&prim[i+dij[num-1]])            {                vis[i]=1;                dij[num]=i;                dfs(num+1,dij);                vis[i]=0;            }}int main(){    int t=1,dij[25];    while(~scanf("%d",&n))    {        memset(dij,0,sizeof(dij));        memset(vis,0,sizeof(vis));        dij[1]=1;        vis[1]=1;        printf("Case %d:\n",t++);        dfs(2,dij);        printf("\n");    }    return 0;}