hdoj 4255 A Famous Grid （素数打表+bfs）

A Famous Grid

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 365    Accepted Submission(s): 148

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

 题意应该很容易理解 就是一个蛇形的素数表里面 求两个合数的最短距离

一开始看错题了 以为是1000*1000 后来老吴说是10000*10000 的素数表格 直接做怕超时 所以也就直接放弃了  比赛完之后原来是100*100的表格

瞬间被晕死 虽然BFS没怎么学 但是这次写完之后也稍微有点提高了 继续努力吧 呵呵

#include <iostream>
#include <fstream>
using namespace std;
#define  max 210
#define  fmax max*max
int p[max][max];
int ss[max][max];
int f[max*max];
bool b[max*max];
int dx[4]= {0, -1, 0, 1};
int dy[4]= {1, 0, -1, 0};
int x1[fmax],y1[fmax];

void init()
{
//    freopen("spiral.in","r",stdin);
    //freopen("out.txt","w",stdout);
    int sum=1;
    int x=max/2,y=max/2,i,j;
    p[x][y]=f[sum++];    
    for ( i=1;i<max;i++)
    {
        for ( j=1;j<=i;j++)
        {
            if (i%2)              
            {
                p[x][++y]=f[sum++];    
                x1[sum-1]=x;
                y1[sum-1]=y;
            }
            if (i%2==0)              
            {
                p[x][--y]=f[sum++];    
                x1[sum-1]=x;
                y1[sum-1]=y;
            }
        }
        for ( j=1;j<=i;j++)
        {         
            if (i%2)
            {
                p[--x][y]=f[sum++];    
                x1[sum-1]=x;
                y1[sum-1]=y;
            }
            if (i%2==0)    
            {
                p[++x][y]=f[sum++];    
                x1[sum-1]=x;
                y1[sum-1]=y;
            }
        }
    }
}

void prim()
{
    f[1]=1;
    for (int i=2;i<max*max;i++)
    {
        f[i]=i;
        if (!b[i]) 
        {
            f[i]=0;
            for (int j=i+i;j<max*max;j=j+i)            
                b[j]=1;            
        }
    }
}
int ans;
int begin,end;
int q[fmax*2];
int d[max][max];

int bfs(int x,int y)
{
    begin=end=0;
    q[end++]=x1[x];
    q[end++]=y1[x];
    memset(d,-1,sizeof(d));
    d[x1[x]][y1[x]]=0;
    while (begin<end)
    {
        int x2=q[begin++];
        int y2=q[begin++];
        if (p[x2][y2]==y)        
            return d[x2][y2];        
        for (int i=0;i<4;i++)
        {
            int x3=x2+dx[i];
            int y3=y2+dy[i];
            if (x3>=0&&y3>=0&&x3<max&&y3<max&&p[x3][y3]&&d[x3][y3]==-1)
            {
                d[x3][y3]=d[x2][y2]+1;
                q[end++]=x3;
                q[end++]=y3;
            }
        }
    }
    return -1;
}

int main()
{
    prim();
    init();    
    int i,xx,yy,j;
/*            for ( i=0;i<max;i++)
    {
    for (j=0;j<max;j++)
    {
    cout<<p[i][j]<<" ";
    }
    cout<<endl;        
    }
    */
    int cas=1;
    while (~scanf("%d%d",&xx,&yy))
    {        
        //printf("Case %d: ",cas++);
        ans=0;
        ans=bfs(xx,yy);
        if (ans==-1)
        {
            printf("Case %d: impossible\n",cas++);
        }
        else printf("Case %d: %d\n",cas++,ans);
    }
    
    return 0;
}