[ACM] 1016 Prime Ring Problem （深度优先搜索）

Prime Ring Problem

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

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

 

Source

Asia 1996, Shanghai (Mainland China)

题外话：

省赛的失利也不能阻挡我前进的脚步，经历过就是一件好事，正如老师所说的，结果谁也没法预测，但这个过程是实实在在的，自己可以在其中收获很多东西。人生中需要经历很多事情，有些事情，只有亲身经历过才会懂。

解题思路：

素数环要求任何相邻的两个数相加的和必须为素数。用DFs，从一年前期末考试两个多小时也没做出来这个题到现在的一次Ac，自己真的进步了。

代码：

#include <iostream>#include <algorithm>#include <string.h>#include <cmath>using namespace std;const int maxn=25;bool visit[maxn];int num[maxn];int n;bool prime(int n){    if(n==1)        return false;    if(n==2)        return true;    if(n%2==0)        return false;    for(int i=3;i<=(int)sqrt(n);i+=2)        if(n%i==0)        return false;    return true;}void dfs(int step){    if(step>n&&prime(num[n]+num[1]))    {        for(int i=1;i<=n-1;i++)            cout<<num[i]<<" ";        cout<<num[n]<<endl;    }    for(int i=2;i<=n;i++)    {        num[step]=i;        if(prime(num[step]+num[step-1])&&!visit[i])//继续向下搜索的条件        {            visit[i]=1;            dfs(step+1);            visit[i]=0;        }    }}int main(){    int c=1;    while(cin>>n)    {        cout<<"Case "<<c++<<":"<<endl;        memset(visit,0,sizeof(visit));        num[1]=1;        dfs(2);        cout<<endl;    }    return 0;}