nyoj 592 spiral grid

spiral grid

时间限制：2000 ms  |  内存限制：65535 KB

难度：4

描述

Xiaod 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. In addition, traveling from a prime number is disallowed, either. 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.

 

输入

Each test case is described by a line of input containing two nonprime integer 1 <=x, y<=10,000.

输出

For each test case, display its case number followed by the length of the shortest path or "impossible" (without quotes) in one line.

样例输入

1 4

9 32

10 12

样例输出

Case 1: 1

Case 2: 7

Case 3: impossible

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
int map[105][105]={0};
int mm=10000,n=100;
int fir,end,loop[105][105];
int prime[11000]={1,1,0};
int sx[]={0,0,1,-1},zy[]={1,-1,0,0};
void Build_map() //建立地图
{
    int i,j,k;
    for(i=1;i<=(n+1)/2;i++)
    {
        for(j=1;j<=n-2*i+2;j++)
            map[i][j+i-1]=mm--;
        for(j=1;j<=n-2*i;j++)
           map[j+i][n-i+1]=mm--;
        for(j=n-i+1;j>=i;j--)
           map[n-i+1][j]=mm--;
        for(j=n-i;j>=i+1;j--)
           map[j][i]=mm--;
    }
}
void printf_map()//打印地图
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=n;j++)
          printf("%d ",map[i][j]);
        printf("\n");
    }

}
void judge_prime()  //筛选法打个素数表
{
    int i,j;
    for(i=2;i<=10000;i++)
    {
        if(prime[i]==0)
        {
            for(j=i+i;j<=10000;j+=i)
              prime[j]=1;
        }
    }
}
struct coordinate
{
    int x;
    int y;
    int step;
}t1;
void find() //找到 fir 的坐标，存在 t1 中
{
    int i,j;
    for(i=1;i<=n;i++)
      for(j=1;j<=n;j++)
      {
          if(map[i][j]==fir)
          {
              t1.x=i;
              t1.y=j;
              loop[i][j]=1;
              t1.step=0;
              return;
          }
      }
}
int bfs()
{
    int i,j,step,x,y;
    queue<coordinate> Q;
    Q.push(t1);
    while(!Q.empty())
    {
        i=Q.front().x;
        j=Q.front().y;
        step=Q.front().step;
        Q.pop();
        if(map[i][j]==end)
          return step;
        for(int a=0;a<4;a++)
        {
            x=i+sx[a];
            y=j+zy[a];
            if(map[x][y]!=0&&prime[map[x][y]]==1&&loop[x][y]==0)
            {
                coordinate t2={x,y,step+1};
                Q.push(t2);
                loop[x][y]=1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,nn=0;
    Build_map();
    judge_prime();
    //printf_map();
    while(~scanf("%d%d",&fir,&end))
    {
        nn++;
        memset(loop,0,sizeof(loop));
        find();
        int ans=bfs();
        if(ans)
            printf("Case %d: %d\n",nn,ans);
        else
            printf("Case %d: impossible\n",nn);
    }
}