[HDU 1016]--Prime Ring Problem(回溯)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

Asia
1996, Shanghai (Mainland China)

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题目大意：有一个整数n，把从1到n的数字无重复的排列成环，且使每相邻两个数（包括首尾）的和都为素数，称为素数环。

　　　　　为了简便起见，我们规定每个素数环都从1开始。有多组测试数据，每组输入一个n(0<n<20)，n=0表示输入结束。

　　　　　输出每组第一行输出对应的Case序号，从1开始。如果存在满足题意叙述的素数环，从小到大输出。

解题思路：回溯的思想就是了，一个dfs搞定，最坑爹的是，每输完一组末尾都要加上换行，我没加结果提交wa,明明是pe,各种改,

　　　　　彻底无爱了，Orz~~~

代码如下：

 #include <iostream>
 #include <cstring>
 using namespace std;

 #define maxn 40
 int vis[], x[], T, n;
 int prime[maxn] = { , ,  };
 void init()
 {
     int i, j;
     for (i = ; i <= maxn; i++){
         if (!prime[i]){
             for (j = ; i*j <= maxn; j++)
                 prime[i*j] = ;
         }
     }
 }

 void dfs(int cur){
     if (cur == n&&!prime[ + x[n - ]]){
         for (int i = ; i < n; i++){
             if (i) cout << ' ';
             cout << x[i];
         }
         cout << endl;
     }
     else for (int i = ; i <= n; i++){
         if (!vis[i] && !prime[i + x[cur - ]]){
             x[cur] = i;
             vis[i] = ;
             dfs(cur + );
             vis[i] = ;
         }
     }
 }

 int main(){
     init();
     while (cin >> n){
         cout << "Case " << ++T << ':' << endl;
         if (n == )
             cout <<  << endl;
         else if (n & )
             cout << endl;
         else{
             memset(vis, , sizeof(vis));
             x[] = ;
             dfs();
         }
         cout << endl;//没加这一句pe来个wa,我也是醉了,各种改,无爱了~~~~
     }
     return ;
 }