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[10010];
int dir[4][2]={0,1,0,-1,1,0,-1,0};
const int maxn=105;
int vst[maxn][maxn];//访问标记
int map[105][105],map1[105][105];
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]!=0 && s.x>=0 && s.x<100 && s.y>=0 && s.y<100) )
    return 1;
    else return 0;
}
int  bfs(int aa,int bb)
{int i,j;
memset(vst,0,sizeof(vst));
    for(i=0;i<=100;i++)
     for(j=0;j<=100;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=0;
        q.push(st);
        vst[st.x][st.y]=1;
        while(!q.empty())
        {
            now=q.front();
            q.pop();
            if(map1[now.x][now.y]==bb)
            {
                return now.step;
            }
            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(check(next,bb))//满足条件；
            {
                //cout<<next.x<<"***"<<next.y<<endl;
                q.push(next);
                vst[next.x][next.y]=1;
            }
            }
        }
        return 0;
}
void fun( )//判断是否是素数
{
    int i,j,k;
    memset(a,0,sizeof(a));
       a[1]=1;
       for(i=2;i<=10010;i++)
         for(j=2;i*j<=10010;j++)
           a[i*j]=1;
}
void fuu()//蛇形填数
{
    int tot,x=0,y=0,n=100;
 memset(map,0,sizeof(map));
 memset(map1,0,sizeof(map1));
    tot=map1[0][0]=10000;
    map[0][0]=1;
    while(tot>1)
    {
        while(y+1<n  && !map1[x][y+1])
        {
              --tot;
            if(a[tot]!=0)
            {
             map[x][y+1]=1;//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+1<n && !map1[x+1][y])
        {
            --tot;
            if(a[tot]!=0)
            {map[x+1][y]=1;}
            map1[++x][y]=tot;
        }
        while(y-1>=0 && !map1[x][y-1])
        {    --tot;
            if(a[tot]!=0)
               {map[x][y-1]=1;}
            map1[x][--y]=tot;
        }
        while(x>0  && !map1[x-1][y])
        {
            --tot;
            if(a[tot]!=0)
            {map[x-1][y]=1;}
            map1[--x][y]=tot;
        }
    }
}
int main()
{
    int m,n,nn=1,k;
    fun();fuu();
    while(cin>>m>>n)
    {
        if(m==n)
       {
           printf("Case %d: 0
",nn++);continue;//相同输入零
       }
       else if(a[n]==0)//如果第二个是素数则输出impossible
       {printf("Case %d: impossible
",nn++);continue;}
                  k=bfs(m,n);
       if(k==0)
       printf("Case %d: impossible
",nn++);
       else printf("Case %d: %d
",nn++,k);
    }
    return 0;
}