文章标题 HDU 1016 ： Prime Ring Problem （DFS）

Prime Ring Problem

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

题意：有一个环，有n个数，从1~n，然后将这n个数填进这个环，然后使得相邻的两个数之和是素数，然后将所有符合情况的输出来，第一个数是1
分析：用DFS枚举，用一个数组flag【n】来标记每个数是否已经填进去了，然后用数组ans【n】来存储被填进去的第i个数字。具体看代码，特别注意，输出结果后还得输出一个空格。。。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
int n; 
int flag[25];
int ans[25];
int judge(int a,int b){//判断a+b是否是素数 
    int sum=a+b;
    for (int i=2;i*i<=sum;i++){
        if (sum%i==0)return 0;
    }
    return 1;
}
void dfs(int index){
    if (index>n&&judge(ans[n],1)){//如果已经放了n个数，且第n个数与1之和是素数说明符合题意 
        for (int i=1;i<=n;i++){//输出序列 
            if (i==1)printf ("%d",ans[i]);
            else printf (" %d",ans[i]);
        }
        printf ("\n");
        return;
    }
    for (int i=2;i<=n;i++){
        if (!flag[i]&&judge(i,ans[index-1])){//如果当前的数还没用过且与前一个数之和是素数则符合情况 
            flag[i]=1;//标记已经用过了 
            ans[index]=i;//存储数字 
            dfs(index+1);//执行下一个数字 
            flag[i]=0;
        }
    }
}
int main ()
{
    int cnt=1;
    while (cin>>n){
        memset(flag,0,sizeof (flag));
        ans[1]=1;
        flag[1]=1;
        printf ("Case %d:\n",cnt++);
        dfs(2);
        printf ("\n");//多一行空格
    }       
    return 0;
}