递归 题目89：Prime ring problem 在递归函数一开始退出

题目89：Prime ring problem

时间限制：2 秒

内存限制：128 兆

特殊判题：否

提交：54

解决：12

题目描述：

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.

输入：

n (1 < n < 17).

输出：

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.

样例输入：

6
8

样例输出：

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

提示：

 用printf打印输出。

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<sstream>
using namespace std;
int prime[10]={3,5,7,11,13,17,19,23,29,31};
int re[17];
bool hash[17];//1~16
bool isprime(int x)
{
    for(int i=0;i<10;i++)
        if(x==prime[i]) return true;
        return false;
}
void dfs(int i,int n)
{
    if(i>2&&!isprime(re[i-1]+re[i-2])) return;
    if(i<=n)
    {
        for(int j=1;j<=n;j++)
            if(!hash[j])
            {
                hash[j]=true; 
            //    if(isprime(re[i-1]+j))
                //{    
                    re[i]=j;
                
            /*    for(int k=1;k<=i;k++)
                    printf("re:%d ",re[k]);
                printf("\n");
                for(k=0;k<=n;k++)
                    if(hash[k]) printf("true ");
                    else printf("false ");
                    printf("\n");
                    */
                    dfs(i+1,n);
                    
                    hash[j]=false; //return;
            //    }
            }
            
    }
    else if(isprime(re[1]+re[n]))
    {
        for(i=1;i<n;i++)
            printf("%d ",re[i]);
        printf("%d\n",re[i]);
        return;
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    int i,j,n,ncase=1;
    while(scanf("%d",&n)!=EOF)
    {
        memset(hash,0,sizeof(hash));
        printf("Case %d:\n",ncase++);
        re[1]=1;hash[1]=true;
        dfs(2,n);
        printf("\n");
        
    }
    
    return 0;
}