HDU 1016 DFS

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49383 Accepted Submission(s): 21783

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

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
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;int Prime[45]={0};int a[20],vis[20];//vis[i]为标记节点是否访问，i为节点代号，数组内容为0或1，1代表未被访问//a[i]以数组形式储存符合条件的数字int m;int zz=1;int DFS(int step){    if(step==m+1&&!Prime[a[m]+a[1]])    {        for(int i=1;i<m;i++)        {            printf("%d ",a[i]);        }        cout<<a[m]<<endl;//输出满足题目所需        return 0;    }    for(int i=2;i<=m;i++)    {        if(!vis[i]&&!Prime[i+a[step-1]])//i没被使用过，同时i与上一个数的和为素数        {            vis[i]=1;            a[step]=i;            DFS(step+1);            vis[i]=0;//回溯作用是此数组多次使用        }    }}int main(){    for(int i=1; i<=40; i++)    {        for(int j=2; j*j<=i; j++)        {            if(i%j==0)            {                Prime[i]++;                break;            }        }    }    a[1]=1;    //int m;    int k=1;    while(scanf("%d",&m)!=EOF)    {        printf("Case %d:\n",k++);        DFS(2);        cout<<endl;    }    return 0;}
最近一直在回顾之前所学，发现自己的知识储备已经所剩无几，最近做题一直没有什么思路，这道题看了半天也没啥思路，幸好参考了前辈的代码，DFS和BFS的运用一直是弱项，这道题蛮经典的，增加了我对DFS的理解，感谢大神博客

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