rqnoj 34 紧急救援

rqnoj 34 紧急救援

BFS。找到终点之后可以立即跳出循环，提高效率，但是因为数据规模不大，我没有这么做，见谅~~
以下是我的代码：

#include < iostream >
#define  maxn 1001
#define  INF 20000007
using   namespace  std;
const   long  xd[] = { - 1 , 1 , 0 , 0 },yd[] = { 0 , 0 , 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 ++ ;
    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 ++ ;
}

long  n,bx,by,ex,ey,d[maxn][maxn];
bool  g[maxn][maxn];
queue q;

int  main()
{
    freopen( " data.in " , " r " ,stdin);
    freopen( " data.out " , " w " ,stdout);
    
    cin >> n;
     for ( long  i = 1 ;i <= n;i ++ )
    {
         for ( long  j = 1 ;j <= n;j ++ )
        {
             char  r;
            cin >> r;
            g[i][j] = (r == ' 0 ' ? false : true );
            d[i][j] = INF;
        }
        getchar();
    }
    cin >> bx >> by;
    cin >> ex >> ey;
     //   Input
    
    Clear(q);
    d[bx][by] = 0 ;
    Push(q,bx,by);
     while ( ! Empty(q))
    {
         long  xx,yy;Pop(q,xx,yy);
         for ( long  k = 0 ;k < 4 ;k ++ )
        {
             long  rx = xx + xd[k],ry = yy + yd[k];
             if (rx >= 1 && rx <= n && ry >= 1 && ry <= n &&! g[rx][ry] && d[rx][ry] >= INF)
            {
                d[rx][ry] = d[xx][yy] + 1 ;
                Push(q,rx,ry);
            }
        }
    }
    
    cout << d[ex][ey] << endl;
    
return   0 ;
}