HDU 1016 Prime Ring Problem(深度优先搜索)

Prime Ring Problem

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

Total Submission(s): 39311 Accepted Submission(s): 17325

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)

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

本宝宝一向是不擅长做深搜题目的，宝宝这次豁出去了，鼓足勇气后，，笨宝宝冲向了本道搜索题。。

下面笨宝宝把代码给大家附上。。。。真的没啥好说的。。

就是一直递推找到符合题目的解。。。

AC代码。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int primejudge(int n)
{
for(int i=2; i<=sqrt(n); i++)
if(n%i==0)
return 0;
return 1;
}

int a[25],n;
int book[25];
int k;
void dfs(int num,int k)
{
if(k==n)
{
if(primejudge(a[0]+a[n-1]))
{
for(int i=0; i<n-1; i++)
{
printf("%d ",a[i]);
}
printf("%d\n",a[n-1]);
}
return;
}
for(int m=2; m<=n; m++)
{
if(book[m]==0)
{
if(primejudge(a[num-1]+m))
{
a[num]=m;
book[m]=1;
dfs(num+1,k+1);
book[m]=0;
}
}
}
return ;
}

int main()
{
int j=0;
while(~scanf("%d",&n))
{
memset(book,0,sizeof(book));
j++;
printf("Case %d:\n",j);
a[0]=1;
book[1]=1;
dfs(1,1);
printf("\n");
}
return 0;
}