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Commit c3b23ba

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rain84yanglbme
andauthored
feat: add solutions to lc problem: No.1466 (doocs#2886)
* feat: add iterative TS solution to lc problem: No.1466 * Update README.md * Update README_EN.md * Update Solution2.ts * Create Solution2.py * Create Solution2.java * Create Solution2.cpp * Create Solution2.go * style: format code and docs with prettier --------- Co-authored-by: Libin YANG <contact@yanglibin.info>
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‎solution/1400-1499/1466.Reorder Routes to Make All Paths Lead to the City Zero/README.md‎

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@@ -230,4 +230,155 @@ impl Solution {
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:BFS
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我们可以使用广度优先搜索的方法,从节点 0ドル$ 出发,搜索其他所有节点,过程中,如果遇到需要变更方向的边,则累加一次变更方向的次数。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是题目中节点的数量。
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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 minReorder(self, n: int, connections: List[List[int]]) -> int:
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g = [[] for _ in range(n)]
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for a, b in connections:
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g[a].append((b, 1))
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g[b].append((a, 0))
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q = deque([0])
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vis = {0}
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ans = 0
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while q:
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a = q.popleft()
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for b, c in g[a]:
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if b not in vis:
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vis.add(b)
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q.append(b)
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ans += c
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return ans
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```
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```java
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class Solution {
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public int minReorder(int n, int[][] connections) {
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List<int[]>[] g = new List[n];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (var e : connections) {
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int a = e[0], b = e[1];
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g[a].add(new int[] {b, 1});
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g[b].add(new int[] {a, 0});
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}
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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boolean[] vis = new boolean[n];
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vis[0] = true;
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int ans = 0;
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while (!q.isEmpty()) {
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int a = q.poll();
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for (var e : g[a]) {
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int b = e[0], c = e[1];
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if (!vis[b]) {
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vis[b] = true;
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q.offer(b);
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ans += c;
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}
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}
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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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```cpp
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class Solution {
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public:
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int minReorder(int n, vector<vector<int>>& connections) {
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vector<pair<int, int>> g[n];
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for (auto& e : connections) {
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int a = e[0], b = e[1];
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g[a].emplace_back(b, 1);
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g[b].emplace_back(a, 0);
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}
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queue<int> q{{0}};
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vector<bool> vis(n);
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vis[0] = true;
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int ans = 0;
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while (q.size()) {
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int a = q.front();
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q.pop();
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for (auto& [b, c] : g[a]) {
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if (!vis[b]) {
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vis[b] = true;
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q.push(b);
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ans += c;
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}
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}
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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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func minReorder(n int, connections [][]int) (ans int) {
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g := make([][][2]int, n)
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for _, e := range connections {
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a, b := e[0], e[1]
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g[a] = append(g[a], [2]int{b, 1})
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g[b] = append(g[b], [2]int{a, 0})
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}
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q := []int{0}
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vis := make([]bool, n)
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vis[0] = true
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for len(q) > 0 {
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a := q[0]
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q = q[1:]
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for _, e := range g[a] {
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b, c := e[0], e[1]
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if !vis[b] {
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vis[b] = true
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q = append(q, b)
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ans += c
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}
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}
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}
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return
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}
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```
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```ts
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function minReorder(n: number, connections: number[][]): number {
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const g: [number, number][][] = Array.from({ length: n }, () => []);
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for (const [a, b] of connections) {
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g[a].push([b, 1]);
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g[b].push([a, 0]);
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}
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const q: number[] = [0];
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const vis = new Set<number>();
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vis.add(0);
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let ans = 0;
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while (q.length) {
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const a = q.pop()!;
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for (const [b, c] of g[a]) {
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if (!vis.has(b)) {
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vis.add(b);
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q.push(b);
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ans += c;
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}
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}
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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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<!-- solution:end -->
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<!-- problem:end -->

‎solution/1400-1499/1466.Reorder Routes to Make All Paths Lead to the City Zero/README_EN.md‎

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@@ -229,4 +229,155 @@ impl Solution {
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2: BFS
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We can use the Breadth-First Search (BFS) method, starting from node 0ドル,ドル to search all other nodes. During the process, if we encounter an edge that requires a change of direction, we increment the count of direction changes.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Where $n$ is the number of nodes in the problem.
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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 minReorder(self, n: int, connections: List[List[int]]) -> int:
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g = [[] for _ in range(n)]
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for a, b in connections:
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g[a].append((b, 1))
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g[b].append((a, 0))
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q = deque([0])
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vis = {0}
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ans = 0
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while q:
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a = q.popleft()
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for b, c in g[a]:
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if b not in vis:
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vis.add(b)
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q.append(b)
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ans += c
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return ans
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```
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```java
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class Solution {
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public int minReorder(int n, int[][] connections) {
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List<int[]>[] g = new List[n];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (var e : connections) {
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int a = e[0], b = e[1];
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g[a].add(new int[] {b, 1});
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g[b].add(new int[] {a, 0});
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}
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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boolean[] vis = new boolean[n];
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vis[0] = true;
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int ans = 0;
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while (!q.isEmpty()) {
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int a = q.poll();
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for (var e : g[a]) {
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int b = e[0], c = e[1];
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if (!vis[b]) {
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vis[b] = true;
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q.offer(b);
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ans += c;
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}
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}
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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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```cpp
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class Solution {
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public:
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int minReorder(int n, vector<vector<int>>& connections) {
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vector<pair<int, int>> g[n];
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for (auto& e : connections) {
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int a = e[0], b = e[1];
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g[a].emplace_back(b, 1);
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g[b].emplace_back(a, 0);
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}
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queue<int> q{{0}};
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vector<bool> vis(n);
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vis[0] = true;
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int ans = 0;
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while (q.size()) {
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int a = q.front();
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q.pop();
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for (auto& [b, c] : g[a]) {
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if (!vis[b]) {
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vis[b] = true;
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q.push(b);
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ans += c;
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}
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}
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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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func minReorder(n int, connections [][]int) (ans int) {
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g := make([][][2]int, n)
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for _, e := range connections {
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a, b := e[0], e[1]
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g[a] = append(g[a], [2]int{b, 1})
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g[b] = append(g[b], [2]int{a, 0})
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}
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q := []int{0}
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vis := make([]bool, n)
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vis[0] = true
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for len(q) > 0 {
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a := q[0]
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q = q[1:]
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for _, e := range g[a] {
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b, c := e[0], e[1]
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if !vis[b] {
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vis[b] = true
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q = append(q, b)
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ans += c
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}
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}
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}
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return
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}
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```
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```ts
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function minReorder(n: number, connections: number[][]): number {
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const g: [number, number][][] = Array.from({ length: n }, () => []);
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for (const [a, b] of connections) {
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g[a].push([b, 1]);
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g[b].push([a, 0]);
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}
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const q: number[] = [0];
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const vis = new Set<number>();
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vis.add(0);
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let ans = 0;
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while (q.length) {
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const a = q.pop()!;
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for (const [b, c] of g[a]) {
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if (!vis.has(b)) {
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vis.add(b);
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q.push(b);
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ans += c;
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}
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}
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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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<!-- solution:end -->
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<!-- problem:end -->
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class Solution {
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public:
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int minReorder(int n, vector<vector<int>>& connections) {
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vector<pair<int, int>> g[n];
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for (auto& e : connections) {
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int a = e[0], b = e[1];
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g[a].emplace_back(b, 1);
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g[b].emplace_back(a, 0);
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}
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queue<int> q{{0}};
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vector<bool> vis(n);
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vis[0] = true;
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int ans = 0;
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while (q.size()) {
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int a = q.front();
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q.pop();
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for (auto& [b, c] : g[a]) {
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if (!vis[b]) {
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vis[b] = true;
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q.push(b);
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ans += c;
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}
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}
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}
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return ans;
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}
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};
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func minReorder(n int, connections [][]int) (ans int) {
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g := make([][][2]int, n)
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for _, e := range connections {
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a, b := e[0], e[1]
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g[a] = append(g[a], [2]int{b, 1})
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g[b] = append(g[b], [2]int{a, 0})
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}
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q := []int{0}
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vis := make([]bool, n)
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vis[0] = true
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for len(q) > 0 {
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a := q[0]
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q = q[1:]
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for _, e := range g[a] {
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b, c := e[0], e[1]
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if !vis[b] {
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vis[b] = true
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q = append(q, b)
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ans += c
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}
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}
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}
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return
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}
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class Solution {
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public int minReorder(int n, int[][] connections) {
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List<int[]>[] g = new List[n];
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Arrays.setAll(g, k -> new ArrayList<>());
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for (var e : connections) {
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int a = e[0], b = e[1];
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g[a].add(new int[] {b, 1});
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g[b].add(new int[] {a, 0});
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}
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Deque<Integer> q = new ArrayDeque<>();
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q.offer(0);
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boolean[] vis = new boolean[n];
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vis[0] = true;
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int ans = 0;
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while (!q.isEmpty()) {
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int a = q.poll();
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for (var e : g[a]) {
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int b = e[0], c = e[1];
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if (!vis[b]) {
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vis[b] = true;
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q.offer(b);
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ans += c;
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}
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}
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}
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return ans;
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}
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}

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