HDU 4255 A Famous Grid

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 798    Accepted Submission(s): 318

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

 

 

Source

Fudan Local Programming Contest 2012

 

题目大意：有螺旋数组成的图，素数不能过，求两个数之间怎么过最短。

这个x，y《=10000，模拟图的时候最好按40000建图

 

  1 #include<iostream>

  2 #include<queue>

  3 #include<cstdio>

  4 #include<cstring>

  5 

  6 using namespace std;

  7 

  8 int X,Y;

  9 int map[210][210],vis[210][210];

 10 int prime[40010];

 11 

 12 void getPrime(){

 13     memset(prime,1,sizeof(prime));

 14     prime[1]=0;

 15     for(int i=2;i<40010;i++)

 16         if(prime[i])

 17             for(int j=i*i;j<40010;j+=i)

 18                 prime[j]=0;

 19 }

 20 

 21 void Graph(){

 22     int c=40000;

 23     int i,x=1,y=200,a=1,b=200;

 24     while(c>0){

 25         for(i=x;i<=y;i++)

 26             map[x][i]=c--;

 27         x++;

 28         for(i=x;i<=y;i++)

 29             map[i][y]=c--;

 30         y--;

 31         for(i=y;i>=a;i--)

 32             map[b][i]=c--;

 33         b--;

 34         for(i=b;i>a;i--)

 35             map[i][a]=c--;

 36         a++;

 37     }

 38 }

 39 

 40 struct node{

 41     int x,y;

 42     int step;

 43 };

 44 

 45 int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};

 46 

 47 int BFS(int si,int sj,int c){

 48     queue<node> q;

 49     while(!q.empty())

 50         q.pop();

 51     node info,tmp;

 52     memset(vis,0,sizeof(vis));

 53     info.x=si,info.y=sj,info.step=c;

 54     vis[si][sj]=1;

 55     if(map[info.x][info.y]==Y)

 56         return info.step;

 57     q.push(info);

 58     while(!q.empty()){

 59         tmp=q.front();

 60         q.pop();

 61         for(int i=0;i<4;i++){

 62             info.x=tmp.x+dir[i][0];

 63             info.y=tmp.y+dir[i][1];

 64             info.step=tmp.step+1;

 65             if(info.x<1 || info.x>200 || info.y<1 || info.y>200 || prime[map[info.x][info.y]])

 66                 continue;

 67             if(map[info.x][info.y]==Y)

 68                 return info.step;

 69             if(!vis[info.x][info.y]){

 70                 vis[info.x][info.y]=1;

 71                 q.push(info);

 72             }

 73         }

 74     }

 75     return -1;

 76 }

 77 

 78 int main(){

 79 

 80     //freopen("input.txt","r",stdin);

 81 

 82     int cases=0;

 83     getPrime();

 84     Graph();

 85     int si,sj;

 86     while(~scanf("%d%d",&X,&Y)){

 87         int flag=1;

 88         for(int i=1;i<=200 && flag;i++)

 89             for(int j=1;j<=200 && flag;j++){

 90                 if(map[i][j]==X){

 91                     si=i;

 92                     sj=j;

 93                     flag=0;

 94                 }

 95             }

 96         int ans=BFS(si,sj,0);

 97         if(ans==-1)

 98             printf("Case %d: impossible\n",++cases);

 99         else

100             printf("Case %d: %d\n",++cases,ans);

101     }

102     return 0;

103 }