UVA - 524 Prime Ring Problem（dfs回溯法）

UVA - 524 Prime Ring Problem

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Description

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

紫书194页原题

题解：输入正整数n，把1—n组成一个环，是相邻的两个整数为素数。输出时从整数1开始，逆时针排列。同一个环恰好输出一次，n (0 < n <= 16)

暴力解决会超时，应用回溯法，深度优先搜索

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int n;
int a[],vis[];

int isp(int n)           //判断是否为素数
{
    if(n<)
        return false;
    for (int i=;i*i<=n; i++)
    {
        if(n % i == )
            return false;
    }
    return true;
}

void dfs(int s)
{
    if(s==n&&isp(a[]+a[n]))  //递归边界。别忘了测试第一个数和最后一个数
    {
        for(int i=; i<n; i++)
            cout<<a[i]<<" ";
        cout<<a[n]<<endl;
    }
    else
    {
        for(int i=; i<=n; i++)
        {
            if(!vis[i]&&isp(i+a[s]))   //如果i没有用过，并且与钱一个数之和为素数
            {
                a[s+]=i;
                vis[i]=;              //标记
                dfs(s+);
                vis[i]=;              //清除标记
            }
        }
    }
}
int main()
{
    int t=;
    while(cin>>n)
    {
        memset(vis,,sizeof(vis));
        a[]=;
        if(t!=) cout<<endl;            //一定注意输出格式
        t++;
        cout<<"Case "<<t<<":"<<endl;
        dfs();
    }
    return ;
}