NYOJ 592 spiral grid

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

//起点可以是素数，终点不可以是素数 
#include
#include
#include
#include
using namespace std;
int map[105][105];
int vis[105][105];
int prime[10005];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
struct point
{
    int x,y,step;
};
void build()
{
    int i,j,k,num;
    num = 10000;
    k = 1;
    while(num>0)
    {
        for(j=k;j<=100-k;j++)
        {
            map[k][j]=num;
            num --;
        }
        for(i=k;i<=100-k;i++)
        {
            map[i][101-k] = num;
            num --;
        }
        for(j=101-k;j>k;j--)
        {
            map[101-k][j] = num;
            num --;
        }
        for(i=101-k;i>k;i--)
        {
            map[i][k] = num;
            num --;
        }
        k++;
    }
}
int is_prime()
{
    int i,j;
    prime[0]=0;  // 0本来不是素数，在这里用于边界
    prime[1]=1;
    for(i=2;i<=10000;i++)
    {
        if(!prime[i])
        for(j=2*i;j<=10000;j=j+i)
        prime[j]=1;
    }
}
int bfs(int a,int b,int num)
{
    point now,next;
    queueq;
    int i,j,k;
    now.x = a;
    now.y = b;
    now.step = 0;
    q.push(now);
    vis[a][b]=1;
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        for(i=0;i<4;i++)
        {
            next.x = now.x+dir[i][0];
            next.y = now.y+dir[i][1];
            next.step = now.step+1;
            if(map[next.x][next.y]==num)
            return next.step;
            if((!vis[next.x][next.y])&&prime[map[next.x][next.y]])
            {
                q.push(next);
                vis[next.x][next.y]=1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,t,s,e,test,a,b;
    is_prime();
    build();
    test = 0;
    while(scanf("%d%d",&s,&e)==2)
    {
        memset(vis,0,sizeof(vis));
        test ++;
        if(!prime[e])
      {
            printf("Case %d: impossible\n",test);
            continue;
        }
        for(i=1;i<=100;i++)
        for(j=1;j<=100;j++)
        {
            if(map[i][j]==s)
            {
                a=i;
                b=j;
                break;
            }
        }
        t = bfs(a,b,e);
        if(t == 0)
        printf("Case %d: impossible\n",test);
        else
        printf("Case %d: %d\n",test,t);
    }
    return 0;
}