nyoj592 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
唉，好久没写搜索了，竟然写了两个晚上，终于AC了；

错误原因：当被找的是素数是，则不能找到，素数孔，能出不能进，也就是说，输入100 3 输出impossible，而输入3 100，则不是如此；

代码如下：

 #include<algorithm>
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<queue>
 using namespace std;
 int a[];
 int dir[][]={,,,-,,,-,};
 const int maxn=;
 int vst[maxn][maxn];//访问标记
 int map[][],map1[][];
 struct state
 {
     int x,y;//坐标位置；
     int step;//搜索统计
 };
 state mm[maxn];
 bool check(state s,int bb)//判断该点是否满足条件
 {
     //cout<<"**"<<endl;
     if( (map1[s.x][s.y]==bb)||(!vst[s.x][s.y] && map[s.x][s.y]!= && s.x>= && s.x< && s.y>= && s.y<) )
     return ;
     else return ;
 }
 int  bfs(int aa,int bb)
 {int i,j;
 memset(vst,,sizeof(vst));
     for(i=;i<=;i++)
      for(j=;j<=;j++)
        if(map1[i][j]==aa)
            {
               goto end;
            }
         end :
        // cout<<i<<j<<endl;
         queue<state>q;
         state now,next,st;
         st.x=i;st.y=j;
         st.step=;
         q.push(st);
         vst[st.x][st.y]=;
         while(!q.empty())
         {
             now=q.front();
             q.pop();
             if(map1[now.x][now.y]==bb)
             {
                 return now.step;
             }
             for(i=;i<;i++)
             {
                 next.x=now.x+dir[i][];
                 next.y=now.y+dir[i][];
             next.step=now.step+;
             if(check(next,bb))//满足条件；
             {
                 //cout<<next.x<<"***"<<next.y<<endl;
                 q.push(next);
                 vst[next.x][next.y]=;
             }
             }
         }
         return ;
 }
 void fun( )//判断是否是素数
 {
     int i,j,k;
     memset(a,,sizeof(a));
        a[]=;
        for(i=;i<=;i++)
          for(j=;i*j<=;j++)
            a[i*j]=;
 }
 void fuu()//蛇形填数
 {
     int tot,x=,y=,n=;
  memset(map,,sizeof(map));
  memset(map1,,sizeof(map1));
     tot=map1[][]=;
     map[][]=;
     while(tot>)
     {
         while(y+<n  && !map1[x][y+])
         {
               --tot;
             if(a[tot]!=)
             {
              map[x][y+]=;//如果此位置不是素数则能走，能走的为1，否则为零；
             }
             map1[x][++y]=tot;//初始化二位数组,填数
         }
         while( x+<n && !map1[x+][y])
         {
             --tot;
             if(a[tot]!=)
             {map[x+][y]=;}
             map1[++x][y]=tot;
         }
         while(y->= && !map1[x][y-])
         {    --tot;
             if(a[tot]!=)
                {map[x][y-]=;}
             map1[x][--y]=tot;
         }
         while(x>  && !map1[x-][y])
         {
             --tot;
             if(a[tot]!=)
             {map[x-][y]=;}
             map1[--x][y]=tot;
         }
     }
 }
 int main()
 {
     int m,n,nn=,k;
     fun();fuu();
     while(cin>>m>>n)
     {
         if(m==n)
        {
            printf("Case %d: 0\n",nn++);continue;//相同输入零
        }
        else if(a[n]==)//如果第二个是素数则输出impossible
        {printf("Case %d: impossible\n",nn++);continue;}
                   k=bfs(m,n);
        if(k==)
        printf("Case %d: impossible\n",nn++);
        else printf("Case %d: %d\n",nn++,k);
     }
     return ;
 }