码风很好，轻松理解：

#include<bits/stdc++.h>
using namespace std;
int n,m,sx,sy,ex,ey,dx[]={0,0,1,-1},dy[]={1,1,0,-1},xx[]={0,0,1,-1,1,-1,1,-1},yy[]={1,-1,0,0,1,-1,-1,1};
char c[20000];
bool v[20000],vis[20000];
int f(int x,int y){
	return x*n+y;
}
struct node{
	int x,y,s;
};
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%s",c+f(i,1));
	while(~scanf("%d%d%d%d",&sx,&sy,&ex,&ey)&&(sx||sy||ex||ey)){
		memset(v,0,sizeof(v));
		memset(vis,0,sizeof(vis));
		v[f(sx,sy)]=1;
		for(int i=0;i<8;i++){
			int nx=sx+xx[i],ny=sy+yy[i];
			while(nx>=1&&nx<=n&&ny>=1&&ny<=m&&c[f(nx,ny)]=='O'){
				v[f(nx,ny)]=1;
				nx+=xx[i];ny+=yy[i];
			}
		}
		queue<node> q;
		q.push({ex,ey,0});
		vis[f(ex,ey)]=1;
		bool flag=0;
		while(!q.empty()){
			node now=q.front();
			q.pop();
			if(v[f(now.x,now.y)]){
				printf("%d",now.s);
				flag=1;
				break;
			}
			for(int i=0;i<4;i++){
				int nx=now.x+dx[i],ny=now.y+dy[i];
				if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&c[f(nx,ny)]=='O'&&!vis[f(nx,ny)]){
					vis[f(nx,ny)]=1;
					q.push({nx,ny,now.s+1});
				}
			}
		}
		if(!flag) printf("Poor Harry");
		printf("\n");
	}
	return 0;
}