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feat: add solutions to lc problem: No.3243 (doocs#3363)
No.3243.Shortest Distance After Road Addition Queries I
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‎solution/0600-0699/0600.Non-negative Integers without Consecutive Ones/README.md‎

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<pre>
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<strong>输入:</strong> n = 5
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<strong>输出:</strong> 5
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<strong>解释:</strong>
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<strong>解释:</strong>
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下面列出范围在 [0, 5] 的非负整数与其对应的二进制表示:
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0 : 0
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1 : 1

‎solution/0600-0699/0600.Non-negative Integers without Consecutive Ones/README_EN.md‎

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3 : 11
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4 : 100
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5 : 101
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Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
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Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule.
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</pre>
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<p><strong class="example">Example 2:</strong></p>

‎solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README.md‎

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<pre>
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<strong>输入:</strong>nums = [1,2,3,4], n = 4, left = 1, right = 5
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<strong>输出:</strong>13
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<strong>输出:</strong>13
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<strong>解释:</strong>所有的子数组和为 1, 3, 6, 10, 2, 5, 9, 3, 7, 4 。将它们升序排序后,我们得到新的数组 [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] 。下标从 le = 1 到 ri = 5 的和为 1 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13 。
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</pre>
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‎solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README_EN.md‎

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<pre>
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<strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 5
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<strong>Output:</strong> 13
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<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13.
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<strong>Output:</strong> 13
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<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13.
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</pre>
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<p><strong class="example">Example 2:</strong></p>

‎solution/3200-3299/3237.Alt and Tab Simulation/README_EN.md‎

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<ul>
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<li>Initial order: <code>[1,2,3]</code></li>
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<li>After the first query: <code>[<u><strong>3</strong></u>,2,1]</code></li>
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<li>After the second query: <code>[<u><strong>3</strong></u>,2,1]</code></li>
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<li>After the first query: <code>[<u><strong>3</strong></u>,1,2]</code></li>
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<li>After the second query: <code>[<u><strong>3</strong></u>,1,2]</code></li>
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<li>After the last query: <code>[<u><strong>2</strong></u>,3,1]</code></li>
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</ul>
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</div>

‎solution/3200-3299/3243.Shortest Distance After Road Addition Queries I/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:BFS
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我们先建立一个有向图 $\textit{g},ドル其中 $\textit{g}[i]$ 表示从城市 $i$ 出发可以到达的城市列表,初始时,每个城市 $i$ 都有一条单向道路通往城市 $i + 1$。
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然后,我们对每个查询 $[u, v],ドル将 $u$ 添加到 $v$ 的出发城市列表中,然后使用 BFS 求出从城市 0ドル$ 到城市 $n - 1$ 的最短路径长度,将结果添加到答案数组中。
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最后返回答案数组即可。
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时间复杂度 $O(q \times (n + q)),ドル空间复杂度 $O(n + q)$。其中 $n$ 和 $q$ 分别为城市数量和查询数量。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def shortestDistanceAfterQueries(
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self, n: int, queries: List[List[int]]
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) -> List[int]:
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def bfs(i: int) -> int:
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q = deque([i])
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vis = [False] * n
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vis[i] = True
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d = 0
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while 1:
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for _ in range(len(q)):
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u = q.popleft()
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if u == n - 1:
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return d
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for v in g[u]:
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if not vis[v]:
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vis[v] = True
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q.append(v)
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d += 1
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g = [[i + 1] for i in range(n - 1)]
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ans = []
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for u, v in queries:
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g[u].append(v)
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ans.append(bfs(0))
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return ans
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```
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#### Java
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```java
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class Solution {
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private List<Integer>[] g;
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private int n;
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public int[] shortestDistanceAfterQueries(int n, int[][] queries) {
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this.n = n;
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g = new List[n];
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Arrays.setAll(g, i -> new ArrayList<>());
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for (int i = 0; i < n - 1; ++i) {
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g[i].add(i + 1);
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}
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int m = queries.length;
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int[] ans = new int[m];
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for (int i = 0; i < m; ++i) {
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int u = queries[i][0], v = queries[i][1];
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g[u].add(v);
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ans[i] = bfs(0);
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}
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return ans;
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}
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private int bfs(int i) {
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(i);
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boolean[] vis = new boolean[n];
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vis[i] = true;
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for (int d = 0;; ++d) {
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for (int k = q.size(); k > 0; --k) {
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int u = q.poll();
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if (u == n - 1) {
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return d;
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}
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for (int v : g[u]) {
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if (!vis[v]) {
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vis[v] = true;
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q.offer(v);
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}
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}
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}
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}
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
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vector<int> g[n];
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for (int i = 0; i < n - 1; ++i) {
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g[i].push_back(i + 1);
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}
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auto bfs = [&](int i) -> int {
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queue<int> q{{i}};
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vector<bool> vis(n);
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vis[i] = true;
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for (int d = 0;; ++d) {
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for (int k = q.size(); k; --k) {
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int u = q.front();
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q.pop();
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if (u == n - 1) {
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return d;
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}
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for (int v : g[u]) {
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if (!vis[v]) {
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vis[v] = true;
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q.push(v);
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}
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}
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}
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}
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};
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vector<int> ans;
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for (const auto& q : queries) {
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g[q[0]].push_back(q[1]);
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ans.push_back(bfs(0));
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func shortestDistanceAfterQueries(n int, queries [][]int) []int {
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g := make([][]int, n)
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for i := range g {
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g[i] = append(g[i], i+1)
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}
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bfs := func(i int) int {
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q := []int{i}
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vis := make([]bool, n)
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vis[i] = true
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for d := 0; ; d++ {
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for k := len(q); k > 0; k-- {
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u := q[0]
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if u == n-1 {
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return d
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}
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q = q[1:]
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for _, v := range g[u] {
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if !vis[v] {
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vis[v] = true
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q = append(q, v)
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}
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}
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}
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}
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}
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ans := make([]int, len(queries))
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for i, q := range queries {
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g[q[0]] = append(g[q[0]], q[1])
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ans[i] = bfs(0)
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}
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return ans
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}
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```
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#### TypeScript
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```ts
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function shortestDistanceAfterQueries(n: number, queries: number[][]): number[] {
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const g: number[][] = Array.from({ length: n }, () => []);
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for (let i = 0; i < n - 1; ++i) {
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g[i].push(i + 1);
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}
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const bfs = (i: number): number => {
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const q: number[] = [i];
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const vis: boolean[] = Array(n).fill(false);
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vis[i] = true;
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for (let d = 0; ; ++d) {
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const nq: number[] = [];
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for (const u of q) {
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if (u === n - 1) {
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return d;
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}
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for (const v of g[u]) {
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if (!vis[v]) {
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vis[v] = true;
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nq.push(v);
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}
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}
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}
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q.splice(0, q.length, ...nq);
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}
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};
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const ans: number[] = [];
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for (const [u, v] of queries) {
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g[u].push(v);
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ans.push(bfs(0));
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}
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return ans;
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}
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```
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<!-- tabs:end -->

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