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

 

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

int X,Y;
int map[210][210],vis[210][210];
int prime[40010];

void getPrime(){
    memset(prime,1,sizeof(prime));
    prime[1]=0;
    for(int i=2;i<40010;i++)
        if(prime[i])
            for(int j=i*i;j<40010;j+=i)
                prime[j]=0;
}

void Graph(){
    int c=40000;
    int i,x=1,y=200,a=1,b=200;
    while(c>0){
        for(i=x;i<=y;i++)
            map[x][i]=c--;
        x++;
        for(i=x;i<=y;i++)
            map[i][y]=c--;
        y--;
        for(i=y;i>=a;i--)
            map[b][i]=c--;
        b--;
        for(i=b;i>a;i--)
            map[i][a]=c--;
        a++;
    }
}

struct node{
    int x,y;
    int step;
};

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

int BFS(int si,int sj,int c){
    queue<node> q;
    while(!q.empty())
        q.pop();
    node info,tmp;
    memset(vis,0,sizeof(vis));
    info.x=si,info.y=sj,info.step=c;
    vis[si][sj]=1;
    if(map[info.x][info.y]==Y)
        return info.step;
    q.push(info);
    while(!q.empty()){
        tmp=q.front();
        q.pop();
        for(int i=0;i<4;i++){
            info.x=tmp.x+dir[i][0];
            info.y=tmp.y+dir[i][1];
            info.step=tmp.step+1;
            if(info.x<1 || info.x>200 || info.y<1 || info.y>200 || prime[map[info.x][info.y]])
                continue;
            if(map[info.x][info.y]==Y)
                return info.step;
            if(!vis[info.x][info.y]){
                vis[info.x][info.y]=1;
                q.push(info);
            }
        }
    }
    return -1;
}

int main(){

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

    int cases=0;
    getPrime();
    Graph();
    int si,sj;
    while(~scanf("%d%d",&X,&Y)){
        int flag=1;
        for(int i=1;i<=200 && flag;i++)
            for(int j=1;j<=200 && flag;j++){
                if(map[i][j]==X){
                    si=i;
                    sj=j;
                    flag=0;
                }
            }
        int ans=BFS(si,sj,0);
        if(ans==-1)
            printf("Case %d: impossible\n",++cases);
        else
            printf("Case %d: %d\n",++cases,ans);
    }
    return 0;
}