标签：hdu1016 prime ring problem dfs acm
6 8
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 <stdio.h>
#include <string.h>
#define MAX 25
bool visited[MAX] ;
int n , ans[MAX] ;
bool prime[2*MAX];
void DFS(int num)
{
if(num == n)
{
if(!prime[ans[num-1]+1])
return ;
for(int i = 0 ; i < n ; ++i)
{
printf("%d",ans[i]);
if(i != n-1)
{
printf(" ");
}
}
printf("\n") ;
return ;
}
else
{
for(int i = 1 ; i <= n ; ++i)
{
if(!visited[i] && prime[i+ans[num-1]])
{
visited[i] = true ;
ans[num] = i;
DFS(num+1);
visited[i] = false ;
}
}
}
}
int main()
{
for(int i = 2 ; i < 51 ; ++i) prime[i] = true ;
prime[0]=prime[1]=false;
prime[2] = true ;
for(int i = 2 ; i < 51; ++i)
{
for(int j = 2 ; j*i < 51 ; ++j)
{
prime[i*j] = false ;
}
}
ans[0] = 1 ;
int c=1;
while(~scanf("%d",&n))
{
memset(visited,0,sizeof(visited));
visited[1] = true ;
printf("Case %d:\n",c++);
DFS(1);
printf("\n");
}
return 0 ;
}hdu 1016 Prime Ring Problem DFS解法 纪念我在杭电的第一百题
标签：hdu1016 prime ring problem dfs acm
原文地址：http://blog.csdn.net/lionel_d/article/details/43765225
