dfs题：C - Prime Ring Problem



题目：



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：这个题目是一个搜索环，当你输入一个数字n时，你必须把1到n之间的数字（包括1和n）进行排序，但数字1的位置是固定的，所有数字排序成为一个环时必须满足条件：相邻两数相加之和必须为素数，且排序的任意数字只能出现一次。

代码如下：

#include<iostream>using namespace std;int sushu[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//素数为1int n,s[20],mark[20];int dfs(int a){if(a==n && sushu[s[a]+1])//首尾两数之和为素数{for(int i=1;i<n;i++)   cout<<s[i]<<" "; cout<<s[n]<<endl;}else{for(int i=2;i<=n;i++){if(sushu[s[a]+i] && mark[i]==0){mark[i]=1;//进行标记s[a+1]=i;//确定下个位置的数字dfs(a+1);//根据数组下标（从第n个数）进行搜索mark[i]=0;// 回溯，撤消访问标志}}}return 0;}int main(){int k;k=s[1]=1;while(cin>>n){   cout<<"Case "<<k++<<":"<<endl;   memset(mark,0,sizeof(mark));   dfs(1);//根据第一个数1进行搜索   cout<<endl;}return 0;}

这个题目用数组枚举出所属范围内的所有数字之和的情况，是素数则为1，否则为0，为了避免出现重复的数字可以用mark数组进行标记，详细解释请看代码中的注释