hdu A Famous Grid

A Famous Grid

Problem Description

Mr. B has recently discovered the grid named "spiral grid".
Construct the grid like the following figure. (The grid is actually infinite. The figure is only a small part of it.)


Considering traveling in it, you are free to any cell containing a composite number or 1, but traveling to any cell containing a prime number is disallowed. You can travel up, down, left or right, but not diagonally. Write a program to find the length of the shortest path between pairs of nonprime numbers, or report it's impossible.

 

 

Input

Each test case is described by a line of input containing two nonprime integer 1 <=x, y<=10,000.

 

 

Output

For each test case, display its case number followed by the length of the shortest path or "impossible" (without quotes) in one line.

 

 

Sample Input

1 4 9 32 10 12

 

 

Sample Output

Case 1: 1 Case 2: 7 Case 3: impossible

AC代码：

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int x[40000],y[40000];
int valid[200+5][200+5];
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int xa,ya,xb,yb;
struct node
{
  int x,y;
  int step;
};
int isprime( int n)
{
  int i;
  if(n<2) return 0;
  for( i=2;i*i<=n;i++)
  {
    if(n%i==0) return 0;
  }
  return 1;
}
int in( int x,int y)
{
 return x>=0&&y>=0&&x<200&&y<200;
}
void prepare( )
{
  int px,py,i,now,j;
  for( now=1,i=1,px=100,py=100;;i+=2)
  {
    for( j=0;j<i;j++)
    {
     x[now]=px;
     y[now]=py;
     valid[px][py]=!isprime(now);
     px++;
     now++;
     if(!in(px,py)) return ;
    }
    for( j=0;j<i;j++)
    {
     x[now]=px;
     y[now]=py;
     valid[px][py]=!isprime(now);
     py++;
     now++;
     if(!in(px,py)) return ;
    }
    for( j=0;j<=i;j++)
    {
     x[now]=px;
     y[now]=py;
     valid[px][py]=!isprime(now);
     px--;
     now++;
     if(!in(px,py)) return ;
    }
    for( j=0;j<=i;j++)
    {
     x[now]=px;
     y[now]=py;
     valid[px][py]=!isprime(now);
     py--;
     now++;
     if(!in(px,py)) return ;
    }
  }
  
}
int bfs( )
{
 int visit[400][400];
 memset(visit,0,sizeof(visit));
  int i;
  queue<node>q;
  node p;
  p.x=xa;
  p.y=ya;
  p.step=0;
  q.push(p);
  
  while(!q.empty( ))
  {
    node px=q.front( );
    q.pop( );
    xa=px.x;
    ya=px.y;
    visit[xa][ya]=1;
    int dis=px.step;
    if(xa==xb&&ya==yb) 
    {
     printf(" %d\n",dis);
     return  1;
     }
    for( i=0;i<4;i++)
    {
      int xx=xa+dx[i],yy=ya+dy[i];
      if(in(xx,yy)&&valid[xx][yy]&&visit[xx][yy]==0)
      {
         node py;
         py.x=xx,py.y=yy,py.step=dis+1;
         visit[xx][yy]=1;
         q.push(py);
              
      }
    }
    
  }
  return 0;
}
int main( )
{
 int a,b;
 int t=1;
 prepare( );
 while(scanf("%d%d",&a,&b)!=EOF)
 {
   xa=x[a];
   ya=y[a];
   xb=x[b];
   yb=y[b];
   printf("Case %d:",t++);
   if(!bfs( )) puts(" impossible");
 }
 return 0;
}

 链接：http://acm.hdu.edu.cn/showproblem.php?pid=4255