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): 19251    Accepted Submission(s): 8599


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

解题思路：

素数打表，因为n最大是20，所以只要打到40

用DFS进行搜索，然后用vis进行标记，搜索过还需要回溯，为了查找下一组数据，本题数据比较水，不需要简直都可以过

最后还要注意格式：两个数之间要有空格，每组数据要有空行

#include<stdio.h>
#include<cstring>
using namespace std;

int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//素数打表，因为n最大是20，所以只要打到40
int vis[21],a[21],N;

void dfs(int num)
{
if(num==N&&prime[a[num-1]+a[0]])
{
for(int i=0;i<num-1;i++)
printf("%d ",a[i]);
printf("%d\n",a[num-1]);
}
else
{
for(int i=2;i<=N;i++)
{
if(vis[i]==0)
{
if(prime[i+a[num-1]])
{
vis[i]=1;
a[num++]=i;
dfs(num);
vis[i]=0;
num--;
}
}
}
}
}

int main()
{
int num=1;
while(scanf("%d",&N)!=EOF)
{
printf("Case %d:\n",num++);
memset(vis,0,sizeof(vis));
a[0]=1;
dfs(1);
printf("\n");
}
return 0;
}