HDU-1016：Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47403    Accepted Submission(s): 20954

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)

题意：

给定一个数n，求1~n的n个自然数组成的环相邻两两相加为素数。

dfs可解，每搜到一个可行值就继续直到找出所有n个数，再回溯找另一种可能。

AC代码：

#include<bits/stdc++.h>
using namespace std;

int a[21],n;
bool vis[21];

int prime(int a){//判断素数 
    if(a<2)
    return 0;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)
        return 0;
    }
    return 1;
}

void dfs(int s){
    if(s==n&&prime(a[1]+a[n])){//满足条件就输出 
        for(int i=1;i<n;i++)
        cout<<a[i]<<" ";
        cout<<a[n]<<endl;
    }
    else{
        for(int i=2;i<=n;i++){
            if(vis[i]==0&&prime(a[s]+i)){
                vis[i]=1;//标为已用 
                a[s+1]=i;//加入环 
                dfs(s+1);
                vis[i]=0;//标记更新，回溯 
            }
        } 
    }
    
}

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