ZOJ 1457 Prime Ring Problem（dfs+剪枝）


Prime Ring Problem

Time Limit: 10 Seconds
Memory Limit: 32768 KB

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>

using namespace std;

int n,test=0;;
int cir[30];
int vis[30];
int prime[50]={0};

void dfs(int k)
{
if(k==n)
{
if(!prime[cir[k]+cir[1]])
return;
for(int i=1;i<n;i++)
printf("%d ",cir[i]);
printf("%d",cir
);
printf("\n");
return;
}

for(int j=2;j<=n;j++)
{
if(!vis[j]&&prime[cir[k]+j])       //对于相邻素数的推断，在这里才干在zoj,AC，放最前面就仅仅能在HDU，AC了，一层递归调用的时间而已啊！
{

vis[j]=1;
cir[k+1]=j;
dfs(k+1);
vis[j]=0;                      //注意递归复原，这预计是最大的亮点了

}

}
}

int main()
{
prime[3]=1;
prime[5]=1;
prime[7]=1;
prime[11]=1;
prime[13]=1;
prime[17]=1;
prime[19]=1;
prime[23]=1;
prime[29]=1;
prime[31]=1;
prime[37]=1;

while(scanf("%d",&n)!=EOF)
{
//memset(cir,0,sizeof cir);
//memset(vis,0,sizeof vis);
for(int i=1;i<=n;i++)
{
vis[i]=0;
}
printf("Case %d:\n",++test);
if(n % 2 == 1)
{
printf("\n");
continue;
}

cir[1]=1;
dfs(1);
printf("\n");
}

return 0;
}