<p>A city&#39;s <strong>skyline</strong> is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return <em>the <strong>skyline</strong> formed by these buildings collectively</em>.</p>

<p>The geometric information of each building is given in the array <code>buildings</code> where <code>buildings[i] = [left<sub>i</sub>, right<sub>i</sub>, height<sub>i</sub>]</code>:</p>

<ul>
	<li><code>left<sub>i</sub></code> is the x coordinate of the left edge of the <code>i<sup>th</sup></code> building.</li>
	<li><code>right<sub>i</sub></code> is the x coordinate of the right edge of the <code>i<sup>th</sup></code> building.</li>
	<li><code>height<sub>i</sub></code> is the height of the <code>i<sup>th</sup></code> building.</li>
</ul>

<p>You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height <code>0</code>.</p>

<p>The <strong>skyline</strong> should be represented as a list of &quot;key points&quot; <strong>sorted by their x-coordinate</strong> in the form <code>[[x<sub>1</sub>,y<sub>1</sub>],[x<sub>2</sub>,y<sub>2</sub>],...]</code>. Each key point is the left endpoint of some horizontal segment in the skyline except the last point in the list, which always has a y-coordinate <code>0</code> and is used to mark the skyline&#39;s termination where the rightmost building ends. Any ground between the leftmost and rightmost buildings should be part of the skyline&#39;s contour.</p>

<p><b>Note:</b> There must be no consecutive horizontal lines of equal height in the output skyline. For instance, <code>[...,[2 3],[4 5],[7 5],[11 5],[12 7],...]</code> is not acceptable; the three lines of height 5 should be merged into one in the final output as such: <code>[...,[2 3],[4 5],[12 7],...]</code></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/01/merged.jpg" style="width: 800px; height: 331px;" />
<pre>
<strong>Input:</strong> buildings = [[2,9,10],[3,7,15],[5,12,12],[15,20,10],[19,24,8]]
<strong>Output:</strong> [[2,10],[3,15],[7,12],[12,0],[15,10],[20,8],[24,0]]
<strong>Explanation:</strong>
Figure A shows the buildings of the input.
Figure B shows the skyline formed by those buildings. The red points in figure B represent the key points in the output list.
</pre>

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

<pre>
<strong>Input:</strong> buildings = [[0,2,3],[2,5,3]]
<strong>Output:</strong> [[0,3],[5,0]]
</pre>

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

<ul>
	<li><code>1 &lt;= buildings.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= left<sub>i</sub> &lt; right<sub>i</sub> &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>1 &lt;= height<sub>i</sub> &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>buildings</code> is sorted by <code>left<sub>i</sub></code> in&nbsp;non-decreasing order.</li>
</ul>

class Solution {
public:
    struct tree {
        int l, r;
        int tag, val;
    } cnt[400004];
    void Tree(int k, int l, int r) {
        cnt[k].l = l, cnt[k].r = r;
        int mid = (l + r) / 2;
        cnt[k].tag = 0;
        cnt[k].val = 0;
        if (l != r) {
            Tree(2 * k, l, mid);
            Tree(2 * k + 1, mid + 1, r);
        }
    }
    void update(int k, int l, int r, int num) {
        if (l <= cnt[k].l && cnt[k].r <= r) {
            cnt[k].tag = max(cnt[k].tag, num);
            cnt[k].val = max(cnt[k].val, num);
            return;
        }
		if (cnt[k].tag != 0) {
			cnt[2 * k].tag = max(cnt[2 * k].tag, cnt[k].tag);
			cnt[2 * k].val = max(cnt[2 * k].val, cnt[2 * k].tag);
			cnt[2 * k + 1].tag = max(cnt[2 * k + 1].tag, cnt[k].tag);
			cnt[2 * k + 1].val = max(cnt[2 * k + 1].val, cnt[2 * k + 1].tag);
			cnt[k].tag = 0;
		}
        int mid = (cnt[k].l + cnt[k].r) / 2;
        if (l <= mid) { update(2 * k, l, r, num); }
        if (r > mid) { update(2 * k + 1, l, r, num); }
		cnt[k].val = max(cnt[2 * k].val, cnt[2 * k + 1].val);
    }
    int query(int k, int l, int r) {
        if (l <= cnt[k].l && cnt[k].r <= r) {
            return cnt[k].val;
        }
		if (cnt[k].tag != 0) {
			cnt[2 * k].tag = max(cnt[2 * k].tag, cnt[k].tag);
			cnt[2 * k].val = max(cnt[2 * k].val, cnt[2 * k].tag);
			cnt[2 * k + 1].tag = max(cnt[2 * k + 1].tag, cnt[k].tag);
			cnt[2 * k + 1].val = max(cnt[2 * k + 1].val, cnt[2 * k + 1].tag);
			cnt[k].tag = 0;
		}
        int ans = 0;
        int mid = (cnt[k].l + cnt[k].r) / 2;
        if (l <= mid) { ans = max(ans, query(2 * k, l, r)); }
        if (r > mid) { ans = max(ans, query(2 * k + 1, l, r)); }
        return ans;
    }
    vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
        set<int> s;
        for (auto t : buildings) {
            s.insert(t[0]);
            s.insert(t[1]);
        }
        int len = 0;
        map<int, int> tmp;
        int cnt[11011], num[11011];
        for (auto t : s) {
            len++;
            tmp[t] = len;
            cnt[len] = t;
        }
        Tree(1, 1, len);
        // for (auto t : v) {
        //     cout << t << ":" << tmp[t] << endl;
        // }
        for (auto t : buildings) {
            int x = tmp[t[0]], y = tmp[t[1]] - 1;
            // cout << x << "~" << y << ":" << t[2] << endl;
            update(1, x, y, t[2]);
        }
        for (int k = 1; k < len; k++) {
            num[k] = query(1, k, k);
            // cout << k << ":" << num[k] << endl;
        }
        vector<vector<int>> ans;
        int i = 1, j;
        while (i < len) {
            j = i;
            while (j < len && num[i] == num[j]) {
                j++;
            }
            // cout << i << "~" << j << ":" << num[j - 1] << endl;
            ans.push_back(vector<int> {cnt[i], num[j - 1]});
            i = j;
        }
        ans.push_back(vector<int> {cnt[len], 0});
        return ans;
    }
};
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