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