[ACM] hdu 1016 Prime Ring Problem （DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23268    Accepted Submission(s): 10363

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

 

 

Source

Asia 1996, Shanghai (Mainland China)

 

解题思路：

本质是dfs生成全排列，只不过本题有约束条件。由上学期期末考试时2个多小时做不出来这道题，到现在很快做出这道题，看来自己真进步了不少。继续加油！

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
int num[22];
int visit[22];
int n;

bool prime(int n)
{
    if(n==1)
        return 0;
    for(int i=2;i<n;i++)
        if(n%i==0)
    {
        return 0;
    }
    return 1;
}

void dfs(int cur)
{
    if(cur>n&&prime(num[1]+num[n]))//输出条件
    {
        for(int i=1;i<=n-1;i++)
            cout<<num[i]<<" ";cout<<num[n];//注意格式，最后一个数后面没有空格
        cout<<endl;
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(prime(i+num[cur-1])&&!visit[i])//注意是num[cur-1]而不是num[i-1]，因为要比较相邻的两个数，num[cur]前面的数是num[cur-1]
            {
                visit[i]=1;
                num[cur]=i;
                dfs(cur+1);
                visit[i]=0;
            }
        }
    }
}

int main()
{
    int c=1;
    while(cin>>n)
    {
        num[1]=1;
        visit[1]=1;
        memset(visit,0,sizeof(visit));
        cout<<"Case "<<c++<<":"<<endl;
        dfs(2);
        cout<<endl;
    }
    return 0;
}

运行：