題目

分析

考慮迷宮規格明示分治，
也就是每次分成左上左下右上右下處理

程式碼

#include <cstdio>
#define rr register
using namespace std;
int n,x,y;
inline void dfs(int lx,int rx,int ly,int ry,int zx,int zy){
	if (lx+1==rx&&ly+1==ry){
		if (zx==lx&&zy==ly) printf("%d %d 1\n",rx,ry);
		if (zx==lx&&zy==ry) printf("%d %d 2\n",rx,ly);
		if (zx==rx&&zy==ly) printf("%d %d 3\n",lx,ry);
		if (zx==rx&&zy==ry) printf("%d %d 4\n",lx,ly);
		return;
	}
	rr int midx=(lx+rx)>>1,midy=(ly+ry)>>1;
	rr int opt=(zx>midx)<<1|(zy>midy);
	switch (opt){
		case 0:{
			dfs(lx,midx,ly,midy,zx,zy);
			printf("%d %d 1\n",midx+1,midy+1);
			dfs(lx,midx,midy+1,ry,midx,midy+1);
			dfs(midx+1,rx,ly,midy,midx+1,midy);
			dfs(midx+1,rx,midy+1,ry,midx+1,midy+1);
			break;
		}
		case 1:{
			dfs(lx,midx,ly,midy,midx,midy);
			dfs(lx,midx,midy+1,ry,zx,zy);
			printf("%d %d 2\n",midx+1,midy);
			dfs(midx+1,rx,ly,midy,midx+1,midy);
			dfs(midx+1,rx,midy+1,ry,midx+1,midy+1);
			break;
		}
		case 2:{
		    dfs(lx,midx,ly,midy,midx,midy);
			dfs(lx,midx,midy+1,ry,midx,midy+1);
			dfs(midx+1,rx,ly,midy,zx,zy);
			printf("%d %d 3\n",midx,midy+1); 
			dfs(midx+1,rx,midy+1,ry,midx+1,midy+1);
			break;
		}
		case 3:{
			dfs(lx,midx,ly,midy,midx,midy);
			dfs(lx,midx,midy+1,ry,midx,midy+1);
			dfs(midx+1,rx,ly,midy,midx+1,midy);
			dfs(midx+1,rx,midy+1,ry,zx,zy);
			printf("%d %d 4\n",midx,midy);
			break;
		}
	}
}
signed main(){
	scanf("%d%d%d",&n,&x,&y),n=1<<n;
	dfs(1,n,1,n,x,y); return 0;
}