HDU.1016 Prime Ring Problem【DFS递归+素数打表法】（3.12）-优快云博客

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38929 Accepted Submission(s): 17184

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

【代码如下】
#include <stdio.h>
#include <iostream>
#include <cstring>//memset
#include<math.h>
#include<algorithm>


using namespace std;
int n, kase = 0;
int op[25]={0}, rightcope[25]={0};
int prime[45] = {0};
void dfs(int num, int pos)//num代表前一个位置，pos代表环正在确定的数字的位置
{
    if(pos == n && !prime[num+1])//已经成环，1是那个起始数字，即判断是否成环，而且最后一位和第一位（也就是1）之和是否是素数
    {
        for(int i=0; i<n; i++)
        {
            if(i != 0) cout<<" ";
            cout<<rightcope[i];
        }//进行输出
        cout<<endl;
    }
    //如果不满足成环
    //继续深搜
    for(int i=1; i<=n; i++)
    {
        if(!op[i] && !prime[num+i])//找出还没有被占用的数字，并且它的前一位和它的和是素数
        {
            op[i] = 1;
            rightcope[pos] = i;
            dfs(i, pos+1);
            op[i] = 0;//回溯，dfs进行结束后进行；
        }
    }
}


int main()
{
    for(int i=2;i<22;i++)//筛选法打表，素数用0标记
        for(int j=i+i;j<45;j+=i)
            prime[j]=1;
    while(scanf("%d", &n)!= EOF)
    {
        printf("Case %d:\n", ++kase);
       // memset(op, 0, sizeof(op));//op[]记录数字是否被占用
        //memset(rightcope, 0, sizeof(rightcope));//rightcope记录环的第一个位置
        op[1] = 1;//数字1被占用
        rightcope[0] = 1;//刚开始环从第一个位置开始
        dfs(1,1);//对下一位应该填什么进行深搜
        cout<<endl;
    }
}