HDU 1016 Prime Ring Problem【DFS】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61907    Accepted Submission(s): 26691

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

题解：dfs从1开始找，循环判断即可。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
using namespace std;
int n,ans[21];
int vis[105];
int k=1;
int is_Prime(int p){//素数判断
    int a=sqrt(p);
    if(p==0||p==1)
        return 0;
    for(int i=2;i<=a;i++){
        if(p%i==0) return 0;
    }
    return 1;
}
void dfs(int x){
    if(k==n+1&&is_Prime(ans[n]+ans[1])){//如果已经找到n个数并且第n个与1的和是素数则足以构成环，输出。
        for(int i=1;i<n;i++){
            printf("%d ",ans[i]);
        }
        printf("%d
",ans[n]);
    }
    for(int i=2;i<=n;i++){
        if(!vis[i]&&is_Prime(x+i)){//判断是否已经在素数序列中并且是否能够和前一个数构成素数
            vis[i]=1;
            ans[k++]=i;
            dfs(i);
            vis[i]=0;//找其他数是否满足条件
            k--;
        }
    }
}
int main()
{
    int cnt=0;
    while(~scanf("%d",&n)&&n){
        k=1;
        cnt++;//案例个数
        printf("Case %d:
",cnt);
        memset(vis,0,sizeof vis);
        ans[k++]=1;//从1开始
        vis[1]=1;
        dfs(1);//从1开始搜
        printf("
");
    }
    return 0;
}