<p>There are several squares being dropped onto the X-axis of a 2D plane.</p>

<p>You are given a 2D integer array <code>positions</code> where <code>positions[i] = [left<sub>i</sub>, sideLength<sub>i</sub>]</code> represents the <code>i<sup>th</sup></code> square with a side length of <code>sideLength<sub>i</sub></code> that is dropped with its left edge aligned with X-coordinate <code>left<sub>i</sub></code>.</p>

<p>Each square is dropped one at a time from a height above any landed squares. It then falls downward (negative Y direction) until it either lands <strong>on the top side of another square</strong> or <strong>on the X-axis</strong>. A square brushing the left/right side of another square does not count as landing on it. Once it lands, it freezes in place and cannot be moved.</p>

<p>After each square is dropped, you must record the <strong>height of the current tallest stack of squares</strong>.</p>

<p>Return <em>an integer array </em><code>ans</code><em> where </em><code>ans[i]</code><em> represents the height described above after dropping the </em><code>i<sup>th</sup></code><em> square</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/28/fallingsq1-plane.jpg" style="width: 500px; height: 505px;" />
<pre>
<strong>Input:</strong> positions = [[1,2],[2,3],[6,1]]
<strong>Output:</strong> [2,5,5]
<strong>Explanation:</strong>
After the first drop, the tallest stack is square 1 with a height of 2.
After the second drop, the tallest stack is squares 1 and 2 with a height of 5.
After the third drop, the tallest stack is still squares 1 and 2 with a height of 5.
Thus, we return an answer of [2, 5, 5].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> positions = [[100,100],[200,100]]
<strong>Output:</strong> [100,100]
<strong>Explanation:</strong>
After the first drop, the tallest stack is square 1 with a height of 100.
After the second drop, the tallest stack is either square 1 or square 2, both with heights of 100.
Thus, we return an answer of [100, 100].
Note that square 2 only brushes the right side of square 1, which does not count as landing on it.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= positions.length &lt;= 1000</code></li>
	<li><code>1 &lt;= left<sub>i</sub> &lt;= 10<sup>8</sup></code></li>
	<li><code>1 &lt;= sideLength<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
</ul>


class Solution {
public:
    int N=(int)1e9;
    struct Node{
        Node* ls,*rs;
        int val,add;
    };
    Node* root=new Node();
    void update(Node*node,int lc,int rc,int l,int r,int v){
        if(l<=lc&&rc<=r){
            node->val = v;
            node->add = v;
            return ;
        }
        pushdown(node);
        int mid = (lc + rc) >> 1;
        if (l <= mid) update(node->ls, lc, mid, l, r, v);
        if (r > mid) update(node->rs, mid + 1, rc, l, r, v);
        pushup(node);
    }
    int query(Node *node, int lc, int rc, int l, int r) {
        if (l <= lc && rc <= r) return node->val;
        pushdown(node);
        int mid = (lc + rc) >> 1, ans = 0;
        if (l <= mid) ans = query(node->ls, lc, mid, l, r);
        if (r > mid) ans = max(ans, query(node->rs, mid + 1, rc, l, r));
        return ans;
    }
    void pushdown(Node* node) {
        if (node->ls == nullptr) node->ls = new Node();
        if (node->rs == nullptr) node->rs = new Node();
        if (node->add == 0) return ;
        node->ls->add = node->add; node->rs->add = node->add;
        node->ls->val = node->add; node->rs->val = node->add;
        node->add = 0;
    }
    void pushup(Node* node) {
        node->val = max(node->ls->val, node->rs->val);
    }
    vector<int> fallingSquares(vector<vector<int>>& positions) {
        int n=positions.size();
        vector<int>ans;
        for (auto & info : positions) {
            int x = info[0], h = info[1], cur = query(root, 0, N, x, x + h - 1);
            update(root, 0, N, x, x + h - 1, cur + h);
            ans.push_back(root->val);
        }
        return ans;
    }
};
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