HDU 4255A Famous Grid(裸BFS)

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1152    Accepted Submission(s): 451

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：上面介绍地很清楚了，按照上面的表，然后从一个合数到达另一个合数需要最小的步数。

  解题思路：估计WZY当时敲题目也敲多了，当时也没搭配好，他写了有一段时间，然后又debug，貌似哪里出现了问题，看他写结构体都写了好几个。后来WA了一发，我仔细看了下题目，发现有bug，题目虽然说最大是10000，但是有的情况可以从外面绕，于是改成100w，爆内存，改成80w，又爆内存，50w，时间在1s+的时候WA了。于是又开大点。。当时因为这个贡献了八次的罚时，我想了下思路，按照这样肯定没问题，但是还是WA,最后还有十几分钟时峰峰说他们只开了1.2w，WZY的程序也不知道哪里出了点bug，打素数表没打到位？还是建立矩形表没建好。没找出来。。。今天自己写了一下，直接1A了。。我估计是他建表的时候哪里应该出了点问题。

 题目地址：A Famous Grid

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#define maxn 20000
using namespace std;
int a,b;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

int pa[1005][1005];  //存放放的数字t
int res[1005][1005];
int p[maxn][2];   //存储大小为t的横纵坐标
bool prim[maxn];
bool visi[maxn];
queue<int> mq;

void built()  //建立矩形表
{
    int ci,cj,i,tmp;
    ci=500,cj=500;
    p[1][0]=ci,p[1][1]=cj;
    pa[ci][cj]=1;
    int s=1;

    while(1)
    {
        for(i=1;i<=s;i++)   //向右
        {
            pa[ci][cj+i]=pa[ci][cj+i-1]+1;
            tmp=pa[ci][cj+i];
            p[tmp][0]=ci,p[tmp][1]=cj+i;
        }
        cj+=s;
        for(i=1;i<=s;i++)   //向上
        {
            pa[ci-i][cj]=pa[ci-i+1][cj]+1;
            tmp=pa[ci-i][cj];
            p[tmp][0]=ci-i,p[tmp][1]=cj;
        }
        ci-=s;
        s++;
        for(i=1;i<=s;i++)
        {
            pa[ci][cj-i]=pa[ci][cj-i+1]+1;
            tmp=pa[ci][cj-i];
            p[tmp][0]=ci,p[tmp][1]=cj-i;
        }
        cj-=s;
        for(i=1;i<=s;i++)
        {
            pa[ci+i][cj]=pa[ci+i-1][cj]+1;
            tmp=pa[ci+i][cj];
            p[tmp][0]=ci+i,p[tmp][1]=cj;
        }
        ci+=s;
        s++;
        if(tmp>=15000) break;
    }
}

void sxprim()  //筛选素数
{
    int i,j;
    memset(prim,true,sizeof(prim));
    prim[1]=false;
    for(i=2;i<=sqrt(maxn+0.5);i++)
    {
        if(prim[i])
        {
              for(j=i*i;j<maxn;j+=i)
                    prim[j]=false;
        }
    }
}

void bfs()
{
    while(!mq.empty())
        mq.pop();
    int i;
    mq.push(a);
    while(!mq.empty())
    {
        int tt=mq.front();
        mq.pop();
        if(tt==b)
        {
            break;
        }
        int tx=p[tt][0],ty=p[tt][1];
        int mx,my,cur;
        for(i=0;i<4;i++)
        {
            mx=tx+dir[i][0];
            my=ty+dir[i][1];
            cur=pa[mx][my];
            if(!prim[cur]&&!visi[cur]&&cur<=14000)
            {
                mq.push(cur);
                visi[cur]=true;
                res[mx][my]=res[tx][ty]+1;
            }
        }
    }
}

void output()
{
    int i,j;
    for(i=495;i<=505;i++)
    {
        for(j=495;j<=505;j++)
        {
            printf("%4d ",pa[i][j]);
        }
        cout<<endl;
    }
    cout<<p[3][0]<<" "<<p[3][1]<<endl;

    for(i=1;i<=100;i++)
        if(prim[i])
            cout<<i<<" ";
    cout<<endl;
}

int main()
{
    built();
    sxprim();
    //output();
    int cas=0;
    while(cin>>a>>b)
    {
        if(a==b)
        {
            printf("Case %d: 0\n",++cas);
            continue;
        }
        memset(visi,false,sizeof(visi));
        memset(res,0,sizeof(res));
        bfs();
        int tx=p[b][0],ty=p[b][1];
        if(res[tx][ty]==0)
            printf("Case %d: impossible\n",++cas);
        else
        printf("Case %d: %d\n",++cas,res[tx][ty]);
    }
    return 0;
}

//953MS