HDOJ P1016 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

  
  

   
   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)

 

Recommend

JGShining

 

基础DFS。。。HDOJ上那个15Ms的提交究竟是怎么做到的。。。。

#include <iostream>
using namespace std;

int a[21]={0},s[21]={0},b[40]={0},aa,ii=1;
void dfs(int);
int main(int argc, char *argv[])
{
	b[2]=1;  b[3]=1;  b[5]=1;  b[7]=1;
	b[11]=1; b[13]=1; b[17]=1; b[19]=1;
	b[23]=1; b[29]=1; b[31]=1; b[37]=1; 	
	while (cin>>aa){
		memset(a,0,sizeof(a));
		cout<<"Case "<<ii<<":\n"; 
		a[1]=1; s[1]=1;
		dfs(1); ii++;
		cout<<endl; 
	}
	return 0;
}

void dfs(int i){
	if((i==aa)&&(b[s[i]+s[1]])) 
		for(int j=1;j<=aa;++j){
			cout<<s[j];
			if (j<aa) cout<<' ';
			else cout<<endl;
		}
	else
		for(int j=2;j<=aa;++j)
			if ((!a[j])&&(b[j+s[i]])){
				a[j]=1; s[i+1]=j;
				dfs(i+1);
				a[j]=0; s[i+1]=0; //s[j]=0; 可以去掉 
			}
}

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