RT,做法与dalao们不同...注释已给出

#include <bits/stdc++.h>
using namespace std;
struct node {
	int x, y, d;
};
int m, n, l[15], x, y, px, py, maxleaf = 1, cnt;
char a[5010][5010];
bool del[5010][5010];
queue <node> q;
inline int leaf (int x) { return maxleaf >> (x - 1); }
//某行第一个节点的叶子数
inline int space (int x) {
	if (x == m) return 0;
	return (leaf(x) >> 1) * 3 - 1;
}
//第x行所需要留出的空格
inline char dif (char x) {
	if (x == '\\') return '/';
	if (x == 'o') return 'o';
	return '\\';
}
//不同的字符
void putc (int x, int y, int xx, int yy) {
	while (x != xx && y != yy) {
		++x, --y;
		a[x][y] = '/';
	}
}
//填充两节点间路径
void copy (int x, int y, int d=-1) {
	if (x == l[m]) return ;
	int tx = x, ty = y, ny;
	while (a[tx][ty]) {
		++tx, ty += d;
		if (d > 0) ny = y-(ty-y);
		else ny = y+(y-ty);
		a[tx][ny] = dif (a[tx][ty]);
		if (a[tx][ty] == 'o')
			q.push ((node) {tx, y+(y-ty), -d});
	}
}
//以(x,y)为对称轴复制
void build () {
	while (!q.empty ()) {
		node fro = q.front ();
		q.pop ();
		int x = fro.x, y = fro.y, d = fro.d;
		copy (x, y, d);
	}
}
//建立满二叉树
void delup (int x, int y) {
	int d;
	a[x][y] = ' ';
	if (a[x-1][y+1] == '/') d = 1;
	if (a[x-1][y-1] == '\\') d = -1;
	while (a[x][y] != 'o') {
		a[x][y] = ' ';
		--x, y += d;
	}
}
//向上删除
void deldown (int x, int y, int d) {
	a[x][y] = ' ';
	if (x == l[m]) return ;
	while (a[x][y] >= ' ') {
		if (a[x][y] == 'o') deldown (x, y, -d);
		a[x][y] = ' ';
		++x, y += d;
	}
}
//向下删除(遇到分支则递归同时处理)
int main () {
	scanf ("%d%d", &m, &n);
	maxleaf = 1 << (m-1);
	for (int i=1, j=1; j<=m; ++j) {
		l[j] = x = i, y = space (j) + 1;
		q.push ((node) {x, y, -1});
		if (j != 1) putc (px, py, x, y);
		a[x][y] = 'o';
		i += (space(j) - space (j+1));
		px = x, py = y;
	}
   //建立一条链,以供copy函数复制为二叉树
	build ();
	for (int i=1; i<=n; ++i) {
		scanf ("%d%d", &x, &y);
		cnt = 0;
		for (int j=1; j<=space (1) * 2+1; ++j) {
			if (a[l[x]][j] == 'o' || del[l[x]][j]) ++cnt;
			if (cnt == y) {
				del[l[x]][j] = true;
				delup (l[x], j);
				deldown (l[x], j, 1);
				deldown (l[x], j, -1);
				break ;
			}
		}
	}
   //删点
	for (int i=1; i<=l[m]; ++i) {
		for (int j=1; j<=space (1) * 2+1; ++j) {
			if (a[i][j] > ' ')
				putchar (a[i][j]);
			else putchar (' ');
		}
		putchar ('\n');
	}
   //输出
}