hdu 1016 Prime Ring Problem（dfs，素数环）

Prime Ring Problem

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

题目大意：

给出一个数 n  ，用 1 - n  组成一个环，这个换的要求是，从 1 开始，并且每两个相邻的数的和都为素数

分析：

这就是典型的素数环，刘汝佳《算法竞赛入门经典》 上有讲解，我在这就不多说了，

每次选一个数，判断这个数和前一个数的和是否为 素数，如果是进行下一个数，否则换数，如果这个数 是第 n 个数，那么需要判断是否与第一个数的和也为素数。这里面判断素数的时候不能够每次都用判断素数的方法判断一遍，那样会超时，用一个数组直接事先打一个素数表就行了

附上代码：

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <queue>#include <algorithm>#include <map> using namespace std;/* *   筛选法素数打表  */const int MAXN = 100;bool is_prime[MAXN];void init(){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 = i * i;j < MAXN;j+=i){is_prime[j] = false;}}}}int visit[100];int n;void DFS(int t,int ans[]){if(t > n)return;if(t == n && is_prime[ans[n - 1] + ans[0]]){for(int i = 0;i < n;i++){if(i != 0)printf(" ");printf("%d",ans[i]);}printf("\n");}for(int i = 1;i <= n;i++){if(!visit[i] && is_prime[ans[t - 1] + i]){ans[t] = i;visit[i] = 1;DFS(t + 1,ans);visit[i] = 0;ans[t] = 0;}}}int main(){init();int flag = 1;while(cin >> n){printf("Case %d:\n",flag++);memset(visit,0,sizeof(visit));int ans[100] = {0};ans[0] = 1;visit[1] = 1;DFS(1,ans);printf("\n");}return 0;}