hdu_1016（剪枝深度优先搜索）

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

#include<iostream>#include<stdio.h>#include<cstring>using namespace std;int a[6];bool prime[40];int n;bool vis[25];void show(){    for(int i=0;i<n;i++){        cout << a[i] << ' ' ;    }    cout << endl;}void getPrime(){    memset(prime,true,sizeof(prime));    prime[1]=false;    for(int i=2;i<=40;i++)        if(prime[i])        {            for(int j=i+i;j<=40;j+=i)                prime[j]=false;        }}void dfs(int cur){    if(cur==n && prime[1+a[n-1]]){        show();        return;    }    for(int i=2;i<=n;i++){        if(vis[i])            continue;        if(prime[a[cur-1]+i]){            a[cur]=i;            vis[i]=true;            dfs(cur+1);            vis[i]=false;        }    }}int main(){    a[0]=1;    getPrime();    cout << "n: " << endl;    while(scanf("%d",&n)==1){        dfs(1);    }    return 0;}