Shortest Distance After Road Addition Queries II
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/3200-3299/3244.Shortest%20Distance%20After%20Road%20Addition%20Queries%20II/images/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>
Greedy; Graph; Array; Ordered Set
func shortestDistanceAfterQueries(n int, queries [][]int) (ans []int) {
	nxt := make([]int, n-1)
	for i := range nxt {
		nxt[i] = i + 1
	}
	cnt := n - 1
	for _, q := range queries {
		u, v := q[0], q[1]
		if nxt[u] > 0 && nxt[u] < v {
			i := nxt[u]
			for i < v {
				cnt--
				nxt[i], i = 0, nxt[i]
			}
			nxt[u] = v
		}
		ans = append(ans, cnt)
	}
	return
}
