题目1459：Prime ring problem

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

Code:

#include <cstdio>
 
using namespace std;
 
const int arrSize=22;
int ans[arrSize];     //保存环中每一个被放入的数
bool hash[arrSize];   //标记之前已经被放入环中的数
int n;
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41};  //在给定范围内的素数集合
 
bool isPrime(int x){
    for(int cnt=0;cnt<13;++cnt){
        if(x==prime[cnt])
            return true;
    }
    return false;
}
 
void check(){                             //检查输出由回溯法枚举得到的解
    if(isPrime(ans[n]+ans[1])==false)     //判断最后一个数与第一个数的和是否为素数，若不是则直接返回
        return;
    for(int cnt=1;cnt<=n;++cnt){
        if(cnt!=1)
            printf(" ");
        printf("%d",ans[cnt]);
    }
    printf("
");
}
 
void DFS(int num){     //递归枚举,num为ans数组中已经放入数字的个数
    if(num>1){         //当放入数字的个数大于1时
        if(isPrime(ans[num]+ans[num-1])==false)
            return;
    }
    if(num==n){        //当放入的数字的个数等于n时
        check();
        return;
    }
    for(int cnt=2;cnt<=n;++cnt){
        if(hash[cnt]==false){
            hash[cnt]=true;
            ans[num+1]=cnt;
            DFS(num+1);
            hash[cnt]=false;   //能执行到这儿就说明当前num+1个数确定为ans[1]到ans[num+1]时,
        }                      //所有情况都已经处理过，所以要把hash[i]重新标记为未使用过。
    }
}
 
int main()
{
    int cas=0;
    while(scanf("%d",&n)!=EOF){
        ++cas;
        for(int cnt=0;cnt<arrSize;++cnt){
            hash[cnt]=false;
        }
        ans[1]=1;
        printf("Case %d:
",cas);
        hash[1]=true;
        DFS(1);
        printf("
");
    }
    return 0;
}
 
/**************************************************************
    Problem: 1459
    User: lcyvino
    Language: C++
    Result: Accepted
    Time:590 ms
    Memory:1020 kb
****************************************************************/