HDU-4255

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1496    Accepted Submission(s): 567

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

/**       

          题意：给出两个数，问两点之间的最短距离

          做法：蛇形矩阵 + bfs + 优先队列

**/

#include <iostream>

#include <stdio.h>

#include <cmath>

#include <algorithm>

#include <string.h>

#include <queue>

#define maxn 40000

using namespace std;

int mmap[200][200];

int a[200][200];

int vis[200][200];

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

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

int n,m;

bool num[maxn];

void is_prime()

{

    int tot = 0;

    memset(num,false,sizeof(num));

    num[1] = true;

    for(long long  i=2; i<maxn; i++)

    {

        if(!num[i])

        {

            for(long long j=i*i; j<=maxn; j+=i)

            {

                num[j] = true;

            }

        }

    }

}

struct Node

{

    int x;

    int y;

    int step;

    Node() {}

    Node(int _x,int _y,int _step)

    {

        x = 0;

        y = 0;

        step =0;

    }

} start,endd;

struct cmp

{

    bool operator () (const Node &a,const Node &b)

    {

        return a.step>b.step;

    }

};

int check(int x,int y)

{

    if(x>=0 && x <200 && y >= 0 && y < 200 && num[a[x][y]] == true&& !vis[x][y]) return 1;

    return 0;

}

priority_queue<Node,vector<Node>,cmp >que;

bool bfs()

{

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

    Node tmp,now;

    while(!que.empty()) que.pop();

    que.push(start);

    vis[start.x][start.y] = 1;

    start.step = 0;

    while(!que.empty())

    {

        now = que.top();

        que.pop();

        //cout<<now.x<<"   "<<now.y<<"   "<<now.step<<endl;

        if(now.x == endd.x && now.y == endd.y)

        {

            endd.step = now.step;

            return true;

        }

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

        {

            tmp.x = now.x + dx[i];

            tmp.y = now.y + dy[i];

            tmp.step = now.step + 1;

            if(check(tmp.x,tmp.y))

            {

                vis[tmp.x][tmp.y] = 1;

                que.push(tmp);

            }

        }

    }

    return false;

}

void init()

{

    int x = 0;

    int y = 0;

    int nn = 200;

    int num=a[0][0]=40000;

    while(num>1)

    {

        while((y+1)<nn&&!a[x][y+1])  a[x][++y]= --num;

        while((x+1)<nn&&!a[x+1][y])  a[++x][y]= --num;

        while((y-1)>=0&&!a[x][y-1]) a[x][--y]= --num;

        while((x-1)>=0&&!a[x-1][y]) a[--x][y]= --num;



    }

}

int main()

{

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

    init();

    is_prime();

    int Case = 1;

    while(~scanf("%d %d",&n,&m))

    {

        for(int i=0; i<200; i++)

        {

            for(int j=0; j<200; j++)

            {

                if(a[i][j] == n)

                {

                    start.x = i;

                    start.y = j;

                }

                if(a[i][j] == m)

                {

                    endd.x= i;

                    endd.y = j;

                }

            }

        }

        bool prime = false;

        prime = bfs();

        printf("Case %d: ",Case++);

        if(prime) printf("%d\n",endd.step);

        else printf("impossible\n");

    }

    return 0;

}