HDU 1016 Prime Ring Problem（dfs）

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

 

用深搜回溯，一个一个搜下去

#include<cstdio>
#include<string.h>
#include<cmath>
#include<algorithm>
using namespace std;
int n,cnt;
int prime(int x){//判断素数 
    if(n==1)
        return 0;
    if(n==2)
        return 1;
    for(int i=2;i<=floor(sqrt(x));i++){
        if(x%i==0)
            return 0;
    }
    return 1;
}

int vis[30],a[30];
void dfs(int x){
    if(x==n&&prime(a[n]+1)){//如果正好轮了n个数，且相邻之和都为素数，输出 
        for(int i=1;i<=n-1;i++){
            printf("%d ",a[i]);
        }    
        printf("%d\n",a[n]);    
    }
    for(int i=2;i<=n;i++){ //从2-n一个一个试下去 
        a[cnt]=i;
        if(prime(a[cnt]+a[cnt-1])&&vis[i]==0){ //然后是素数且没有用过就继续 
            vis[i]=1;//标记用过 
            cnt++;
            dfs(x+1);
            vis[i]=0;//回溯 
            cnt--;
        }
    }
}

int main(){
    int q=1;
    while(scanf("%d",&n)!=EOF){
        printf("Case %d:\n",q++);
        memset(vis,0,sizeof(vis));
        memset(a,0,sizeof(0));
        vis[1]=1;
        a[1]=1;
        cnt=2;
        dfs(1);        
        printf("\n");
    }
    return 0; 
}