UVA-524 Prime Ring Problem

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16)

Output T

he 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.

You are to write a program that completes above process.

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

分析：

dfs回溯，判断两数和是否为素数以及是否重复使用作为剪枝条件。

Source：

#include<stdio.h>
#include<string.h>
int vis[20];
int ans[20];
int n;
int is_prim(int x)         /*判断是否为素数*/
{
int i;
if(x<=2)               /*首尾两位不能同时为1*/
return 0;
for(i=2;i<x;i++)
if(x%i==0)
{
return 0;
break;
}
if(i==x)
return 1;
}
void dfs(int cur)            /*回溯*/
{
int i;
if(cur==n&&is_prim(1+ans[n-1]))
{
for(i=0;i<n-1;i++)
printf("%d ",ans[i]);
printf("%d\n",ans[n-1]);        /*注意输出格式*/
}
else
{
for(i=2;i<=n;i++)
{
if(!vis[i]&&is_prim(i+ans[cur-1]))     /*剪枝条件*/
{
vis[i]=1;          /*标记*/
ans[cur]=i;
dfs(cur+1);        /*搜索下一位*/
vis[i]=0;          /*取消标记*/
}
}
}
}
int main()
{
int t=0;
while(scanf("%d",&n)!=EOF)
{
t++;
if(t>=2)                     /*注意输出格式*/
printf("\n");
memset(vis,0,sizeof(vis));
ans[0]=1;
printf("Case %d:\n",t);
dfs(1);                        /*注意搜索起点*/
}
return 0;
}