P1185 绘制二叉树

这题要推几个公式，比较麻烦
m层的二叉树需要的最大宽度和高度：
(w = 2^{m - 1} * 3 - 1, m = 2, 3, ...; w = 1, m = 1)
(h = 2^{m - 2} * 3, m = 2, 3, ...; w = 1, m = 1)

#include<iostream>
using namespace std;

const int N = 2048;

char g[N][N];
int st[N][N];
int m, n;

void dfs(int m, int x, int y, int level, int cnt){
    if(st[level][cnt] == 0) g[x][y] = 'o';
    else return;
    if(m == 1) return;
    
    int h;
    if(m >= 3) h = 3 * (1 << (m - 3)) - 1; // 上一阶二叉树的高度 - 1
    else h = 1;
	
    int nlevel = level + 1, ncnt = cnt << 1;
	
    int x1 = x, y1 = y, x2 = x, y2 = y;
    while(h --){ // 画两条边
      if(st[nlevel][ncnt - 1] == 0){
            ++ x1, -- y1;
            g[x1][y1] = '/';
      }
      if(st[nlevel][ncnt] == 0){
            ++ x2, ++ y2;
            g[x2][y2] = '\';
      }   
    }
    ++ x1, --y1, ++ x2, ++ y2;
    
    // 画左右子树
    if(st[nlevel][ncnt - 1] == 0) dfs(m - 1, x1, y1, nlevel, ncnt - 1);
    if(st[nlevel][ncnt] == 0) dfs(m - 1, x2, y2, nlevel, ncnt);
}

int main(){
    cin >> m >> n;
    
    while(n --){
    	int a, b;
    	cin >> a >> b;
    	
    	st[a][b] = 1;
    }
    
    int x = 1, y = (m == 1 ? 1 : 3 * (1 << (m - 2)));
    
    dfs(m, x, y, 1, 1);
    
    int h = (m == 1 ? 1 : 3 * (1 << (m - 2)));
    int w = (m == 1 ? 1 : (1 << (m - 1)) * 3 - 1);
    
    for(int i = 1; i <= h; i ++){
    	for(int j = 1; j <= w; j ++)
    	    if(g[i][j]) cout << g[i][j];
	    else cout << ' ';
	cout << endl;
    }
	
    return 0;
}