<p>You are given an integer <code>n</code> and a 2D integer array <code>queries</code>.</p>

<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code>. Initially, there is a <strong>unidirectional</strong> road from city <code>i</code> to city <code>i + 1</code> for all <code>0 &lt;= i &lt; n - 1</code>.</p>

<p><code>queries[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents the addition of a new <strong>unidirectional</strong> road from city <code>u<sub>i</sub></code> to city <code>v<sub>i</sub></code>. After each query, you need to find the <strong>length</strong> of the <strong>shortest path</strong> from city <code>0</code> to city <code>n - 1</code>.</p>

<p>There are no two queries such that <code>queries[i][0] &lt; queries[j][0] &lt; queries[i][1] &lt; queries[j][1]</code>.</p>

<p>Return an array <code>answer</code> where for each <code>i</code> in the range <code>[0, queries.length - 1]</code>, <code>answer[i]</code> is the <em>length of the shortest path</em> from city <code>0</code> to city <code>n - 1</code> after processing the <strong>first </strong><code>i + 1</code> queries.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 5, queries = [[2,4],[0,2],[0,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[3,2,1]</span></p>

<p><strong>Explanation: </strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image8.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image9.jpg" style="width: 350px; height: 60px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image10.jpg" style="width: 350px; height: 96px;" /></p>

<p>After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">n = 4, queries = [[0,3],[0,2]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,1]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image11.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.</p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/28/image12.jpg" style="width: 300px; height: 70px;" /></p>

<p>After the addition of the road from 0 to 2, the length of the shortest path remains 1.</p>
</div>

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

<ul>
	<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= queries[i][0] &lt; queries[i][1] &lt; n</code></li>
	<li><code>1 &lt; queries[i][1] - queries[i][0]</code></li>
	<li>There are no repeated roads among the queries.</li>
	<li>There are no two queries such that <code>i != j</code> and <code>queries[i][0] &lt; queries[j][0] &lt; queries[i][1] &lt; queries[j][1]</code>.</li>
</ul>


class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        seg = SegmentTree(n)
        ans = []
        for x, y in queries:
            if x + 1 <= y - 1:
                seg.update(1, 0, n-1, x+1, y-1, 1)
            ans.append(n - seg.tree[1] - 1)

        return ans


class SegmentTree:

    def __init__(self, n):
        self.tree = [0]*(n*4)
        self.lazy = [0]*(n*4)

    def op(self, a, b):
        """
        任意符合結合律的運算
        """
        return a+b

    def push_down(self, id, L, R, M):
        """
        將區間懶標加到答案中
        下推懶標記給左右子樹
        """
        if self.lazy[id]:
            self.tree[id*2] = (M-L+1)
            self.lazy[id*2] = 1
            self.tree[id*2+1] = (R-M)
            self.lazy[id*2+1] = 1
            self.lazy[id] = 0

    def push_up(self, id):
        """
        以左右節點更新當前節點值
        """
        self.tree[id] = self.op(self.tree[id*2], self.tree[id*2+1])

    def update(self, id, L, R, i, j, val):
        """
        區間更新
        對[i, j]每個索引都變成 1
        """
        if i <= L and R <= j:  # 當前區間目標範圍包含
            self.tree[id] = (R - L + 1)
            self.lazy[id] = 1
            return
        M = (L+R)//2
        self.push_down(id, L, R, M)
        if i <= M:
            self.update(id*2, L, M, i, j, val)
        if M < j:
            self.update(id*2+1, M+1, R, i, j, val)
        self.push_up(id)
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