UVa 524 Prime Ring Problem（回溯法）

传送门

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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 ≤ 16)

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. You are to write a program that completes above process.

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

思路

题意：

输入正整数n，把整数1，2，3，……，n组成一个环，使得相邻两个整数之和均为素数。输出时序列头为1开始的序列，同一个环应恰好输出一次

题解:

+ View Code?

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#include<bits/stdc++.h> using namespace std; const int maxn = 40; bool is_prime[maxn],vis[maxn]; int n,a[maxn];   void dfs(int cur) {     if (cur == n && is_prime[a[0] + a[n-1]])  //递归边界，因为是环，所以还要测试第一个和最后一个     {         printf("%d",a[0]);         for (int i = 1;i < n;i++)    printf(" %d",a[i]);         printf("\n");     }     else     {         for (int i = 2;i <= n;i++)    //尝试放置每个数i         {             if (!vis[i] && is_prime[i + a[cur-1]])  //如果i没有用过，并且与前一个数之和为素数             {                 a[cur] = i;                 vis[i] = true;                     //设置使用标志                 dfs(cur+1);                 vis[i] = false;                    //清楚标志             }         }     } }   int main() {     //提前预处理素数表     memset(is_prime,true,sizeof(is_prime));     is_prime[0] = is_prime[1] = false;     for (int i = 2;i < maxn;i++)     {         if (is_prime[i])         {             for (int j = 2 * i;j < maxn;j += i)             {                 is_prime[j] = false;             }         }     }     int tcase = 0;     while (~scanf("%d",&n))     {         memset(vis,false,sizeof(vis));         memset(a,0,sizeof(a));         if (tcase)  printf("\n");         printf("Case %d:\n",++tcase);         a[0] = 1;         dfs(1);     }     return 0; }