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): 52000    Accepted Submission(s): 23028

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

 

一道经典的深搜题，题意应该很容易看懂吧.........

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,num[21],book[21];
//最大输入不超过20，也就是说能达到的最大的素数也就是19+18=37
int prime[12]={2,3,5,7,11,13,17,19,23,29,31,37};
bool isprime(int x){
	for(int i=0;i<12;i++)
		if(x==prime[i])
			return true;
	return false;
}
void dfs(int step){
	//记得要判断第一个点和最后一个点的和是不是素数
	if(step==n&&isprime(1+num[step])){
		for(int i=1;i<=n-1;i++) printf("%d ",num[i]);
		printf("%d\n",num[n]);
	}
	else{
		for(int i=2;i<=n;i++){
			if(!book[i]&&isprime(i+num[step])){
				num[step+1]=i;
				book[i]=1;
				dfs(step+1);
				book[i]=0;
			}
		}
	}
}
int main(){
	int count=1;
	while(~scanf("%d",&n)){
		num[1]=1;
		memset(book,0,sizeof(book));
		printf("Case %d:\n",count++);
		dfs(1);
		cout<<endl;
	}
	return 0;
}