tyvj P1030 乳草的入侵

tyvj P1030 乳草的入侵

Easy BFS。
Here is my code:

#include < iostream >
#define  maxn 107
using   namespace  std;
const   long  xd[] = { - 1 , - 1 , 0 , 1 , 1 , 1 , 0 , - 1 },yd[] = { 0 , - 1 , - 1 , - 1 , 0 , 1 , 1 , 1 };

typedef  struct
{
     long  counter,front,rear,x[maxn * maxn],y[maxn * maxn];
}queue;
void  Clear(queue  & q)
{
    q.counter = 0 ;q.front = 0 ;q.rear =- 1 ;
}
bool  Empty(queue  & q)
{
     return  (q.counter == 0 );
}
void  Push(queue  & q, long  rx, long  ry)
{
    q.counter ++ ;q.rear ++ ;
     if (q.rear >= maxn * maxn) q.rear = 0 ;
    q.x[q.rear] = rx;q.y[q.rear] = ry;
}
void  Pop(queue  & q, long   & rx, long   & ry)
{
    q.counter -- ;rx = q.x[q.front];ry = q.y[q.front];
    q.front ++ ;
     if (q.front >= maxn * maxn) q.front = 0 ;
}

long  n,m,mx,my,ans,g[maxn][maxn],d[maxn][maxn];
queue q;

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */
    cin >> n >> m >> mx >> my;
     for ( long  i = 1 ;i <= m;i ++ )
    {
         char  ch;
         for ( long  j = 1 ;j <= n;j ++ )
        {
            cin >> ch;
             if (ch == ' . ' ) g[i][j] = 1 ;
             else  g[i][j] = 0 ;
        }
    }
     long  t = my;my = mx;mx = m - t + 1 ;
     //   Input
     /*
    for(long i=1;i<=m;i++)
    {
        for(long j=1;j<=n;j++)
            cout<<g[i][j];
        cout<<endl;
    }
    cout<<endl<<mx<<" "<<my<<endl<<endl;
    // */
    
    Clear(q);Push(q,mx,my);
    d[mx][my] = 0 ;g[mx][my] = 0 ;
     while ( ! Empty(q))
    {
         long  xx,yy,nx,ny;Pop(q,xx,yy);
         for ( long  k = 0 ;k < 8 ;k ++ )
        {
            nx = xx + xd[k];
            ny = yy + yd[k];
             if (nx >= 1 && nx <= m && ny >= 1 && ny <= n && g[nx][ny])
            {
                Push(q,nx,ny);
                d[nx][ny] = d[xx][yy] + 1 ;g[nx][ny] = 0 ;
            }
        }
    }
    
    ans = 0 ;
     for ( long  i = 1 ;i <= m;i ++ )
         for ( long  j = 1 ;j <= n;j ++ )
             if (ans < d[i][j])
                ans = d[i][j];
    cout << ans << endl;
    
return   0 ;
}