hdu 4255 BFS 小水

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 579    Accepted Submission(s): 225

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
   
   
   
   

 

题意：按图中方式 打印出map（其实就是蛇形填数） 填好数以后如果是素数 则不能走  否则能走 

问从 某点到某点的最短距离 

另外 打表不能为100*100  要大于100  因为它可以从100之外过去 仅仅只是提问的数据最大为100、

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
int a[200][200],s_x,s_y,e_x,e_y,used[200][200];
int prime[40000+100];
struct haha
{
    int x;
    int y;
    int steps;
    friend bool operator<(struct haha a,struct haha b)
    {
        return a.steps>b.steps;
    }
}q,temp;
int step[4][2]={1,0,-1,0,0,1,0,-1};
void get_prime()
{
    int m=(int)sqrt(40000+0.5);
    int c=0,i,j,n=40000;
    memset(prime,0,sizeof(prime));
    for(i=2;i<=m;i++) if(!prime[i])//目前prime 里存储的还不是素数  只是借用这个数组
    {
        for(j=i*i;j<=n;j+=i) 
            prime[j]=1;
    }
    for(i=2;i<=n;i++) prime[i]=!prime[i]; //现在才是素数
}
void get_map()//蛇形填数 
{
    int n=120,x=0,y=0,tot=n*n;
    memset(a,0,sizeof(a));
    a[0][0]=n*n;
    while(tot>1)
    {
        while(y+1<n&&!a[x][y+1]) a[x][++y]=--tot;
        while(x+1<n&&!a[x+1][y]) a[++x][y]=--tot;
        
        while(x-1>=0&&!a[x-1][y]) a[--x][y]=--tot;
        while(y-1>=0&&!a[x][y-1]) a[x][--y]=--tot;
    }
}
int BFS()
{
      int i,xx,yy,n=200;
      priority_queue<struct haha>que;
      memset(used,0,sizeof(used));
      q.x=s_x;
      q.y=s_y;
      q.steps=0;
      used[s_x][s_y]=1;
      que.push(q);
      while(!que.empty())
      {
          
          temp=que.top();
          que.pop();
          if(temp.x==e_x&&temp.y==e_y) return temp.steps;
          for(i=0;i<4;i++)
          {
              xx=temp.x+step[i][0];
              yy=temp.y+step[i][1];
              if(xx>=0&&xx<n&&yy>=0&&yy<n&&!prime[a[xx][yy]]&&!used[xx][yy])
              {
                  q.x=xx;
                  q.y=yy;
                  q.steps=temp.steps+1;
                  que.push(q);
                  used[xx][yy]=1;
              }
          }
      }
      return -1;

}
int main()
{
     int i,j,cas=0,s,e,ans;
     get_prime();
     get_map();
     while(scanf("%d %d",&s,&e)!=EOF)
     {
         for(i=0;i<200;i++)
             for(j=0;j<200;j++)
             {
                 if(a[i][j]==s) {s_x=i;s_y=j;}
                 if(a[i][j]==e) {e_x=i;e_y=j;}
             }
             if(prime[a[s_x][s_y]]||prime[a[e_x][e_y]])
             {printf("Case %d: %d\n",++cas,ans);continue;}
          ans=BFS();
          if(ans==-1)
              printf("Case %d: impossible\n",++cas);
          else printf("Case %d: %d\n",++cas,ans);
     }
     return 0;
}