HDU-1016Prime Ring Problem

 Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

 

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1016

 

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

 

总结 

   昨天学了DFS，今天自己做了一道题，从6点半到8点半，在提交时我还以为不会过，毕竟除了A+B的题，我没一次性AC过一道题，在测试最大数据时，黑框跑了34秒还没停下来，我以为我要超时了，但是并没有，好高兴。网上搜了一下，发现自己的代码还可以优化，可以将素数打表，每次去查表，不用循环，应该能节省一大堆时间。

#include<iostream>
#include<cstdio>
#include<memory.h>
using namespace std;
int N;
int temp[25],cir[25];

int is_prime(int x)  //判断是否为素数
{
    for(int i=2;i*i<=x;i++)
        if(x%i==0) return 0;
    return 1;
}

void dfs(int a,int n)  //深度搜索，查找序列
{
    int a1;
    if(n==N&&is_prime(a+1)){  //停止搜索条件
        for(int i=1;i<N;i++)
            cout <<cir[i]<<' ';
        cout <<cir[N]<<endl;
        return;
    }
    for(int i=2;i<=N;i++){
        a1=i;
        if(temp[a1]==1) continue;
        if(!temp[a1]&&a1>=1&&a1<=N&&is_prime(a1+a)){
            cir[n+1]=a1;
            temp[a1]=1;
            dfs(a1,n+1);
            temp[a1]=0;
        }
    }
}

int main()
{
    int count1=1;
    while(scanf("%d",&N)==1&&N){
        cout <<"Case "<<count1++<<":"<<endl;
        memset(temp,0,sizeof(temp));
        memset(cir,0,sizeof(cir));
        temp[1]=1;
        cir[1]=1;
        dfs(1,1);
        cout <<endl;
    }
}