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‎Problemset/1129-Shortest Path with Alternating Colors/README.md

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# 1129. Shortest Path with Alternating Colors
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You are given an integer `n`, the number of nodes in a directed graph where the nodes are labeled from `0` to `n - 1`. Each edge is red or blue in this graph, and there could be self-edges and parallel edges.
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You are given two arrays `redEdges` and `blueEdges` where:
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* <code>redEdges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is a directed red edge from node <code>a<sub>i</sub></code> to node <code>b<sub>i</sub></code> in the graph, and
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* <code>blueEdges[j] = [u<sub>j</sub>, v<sub>j</sub>]</code> indicates that there is a directed blue edge from node <code>u<sub>j</sub></code> to node <code>v<sub>j</sub></code> in the graph.
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Return an array `answer` of length `n`, where each `answer[x]` is the length of the shortest path from node `0` to node `x` such that the edge colors alternate along the path, or `-1` if such a path does not exist.
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#### Example 1:
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<pre>
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<strong>Input:</strong> n = 3, redEdges = [[0,1],[1,2]], blueEdges = []
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<strong>Output:</strong> [0,1,-1]
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</pre>
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#### Example 2:
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<pre>
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<strong>Input:</strong> n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]
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<strong>Output:</strong> [0,1,-1]
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</pre>
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#### Constraints:
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* `1 <= n <= 100`
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* `0 <= redEdges.length, blueEdges.length <= 400`
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* `redEdges[i].length == blueEdges[j].length == 2`
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* <code>0 <= a<sub>i</sub>, b<sub>i</sub>, u<sub>j</sub>, v<sub>j</sub> < n</code>
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## Solutions (Rust)
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### 1. Solution
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```Rust
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use std::cmp::Reverse;
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use std::collections::BinaryHeap;
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use std::collections::HashSet;
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impl Solution {
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pub fn shortest_alternating_paths(
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n: i32,
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red_edges: Vec<Vec<i32>>,
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blue_edges: Vec<Vec<i32>>,
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) -> Vec<i32> {
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let n = n as usize;
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let mut red_nexts = vec![vec![]; n];
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let mut blue_nexts = vec![vec![]; n];
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let mut nodes_heap = BinaryHeap::from([(Reverse(0), 0, true), (Reverse(0), 0, false)]);
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let mut visited = HashSet::new();
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let mut answer = vec![(i32::MAX, i32::MAX); n];
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answer[0] = (0, 0);
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for edge in &red_edges {
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red_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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for edge in &blue_edges {
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blue_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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while let Some((Reverse(length), node, is_red)) = nodes_heap.pop() {
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if visited.contains(&(node, is_red)) {
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continue;
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}
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visited.insert((node, is_red));
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if is_red {
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for &next in &blue_nexts[node] {
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if answer[next].1 > length + 1 {
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answer[next].1 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, false));
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}
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}
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} else {
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for &next in &red_nexts[node] {
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if answer[next].0 > length + 1 {
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answer[next].0 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, true));
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}
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}
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}
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}
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answer
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.into_iter()
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.map(|(red, blue)| {
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if red.min(blue) == i32::MAX {
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-1
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} else {
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red.min(blue)
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}
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})
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.collect()
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}
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}
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```

‎Problemset/1129-Shortest Path with Alternating Colors/README_CN.md

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# 1129. 颜色交替的最短路径
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给定一个整数 `n`，即有向图中的节点数，其中节点标记为 `0` 到 `n - 1`。图中的每条边为红色或者蓝色，并且可能存在自环或平行边。
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给定两个数组 `redEdges` 和 `blueEdges`，其中：
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* <code>redEdges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示图中存在一条从节点 <code>a<sub>i</sub></code> 到节点 <code>b<sub>i</sub></code> 的红色有向边，
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* <code>blueEdges[j] = [u<sub>j</sub>, v<sub>j</sub>]</code> 表示图中存在一条从节点 <code>u<sub>j</sub></code> 到节点 <code>v<sub>j</sub></code> 的蓝色有向边。
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返回长度为 `n` 的数组 `answer`，其中 `answer[X]` 是从节点 `0` 到节点 `X` 的红色边和蓝色边交替出现的最短路径的长度。如果不存在这样的路径，那么 `answer[x] = -1`。
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#### 示例 1:
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<pre>
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<strong>输入:</strong> n = 3, redEdges = [[0,1],[1,2]], blueEdges = []
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<strong>输出:</strong> [0,1,-1]
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</pre>
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#### 示例 2:
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<pre>
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<strong>输入:</strong> n = 3, redEdges = [[0,1]], blueEdges = [[2,1]]
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<strong>输出:</strong> [0,1,-1]
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</pre>
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#### 提示:
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* `1 <= n <= 100`
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* `0 <= redEdges.length, blueEdges.length <= 400`
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* `redEdges[i].length == blueEdges[j].length == 2`
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* <code>0 <= a<sub>i</sub>, b<sub>i</sub>, u<sub>j</sub>, v<sub>j</sub> < n</code>
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## 题解 (Rust)
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### 1. 题解
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```Rust
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use std::cmp::Reverse;
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use std::collections::BinaryHeap;
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use std::collections::HashSet;
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impl Solution {
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pub fn shortest_alternating_paths(
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n: i32,
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red_edges: Vec<Vec<i32>>,
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blue_edges: Vec<Vec<i32>>,
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) -> Vec<i32> {
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let n = n as usize;
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let mut red_nexts = vec![vec![]; n];
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let mut blue_nexts = vec![vec![]; n];
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let mut nodes_heap = BinaryHeap::from([(Reverse(0), 0, true), (Reverse(0), 0, false)]);
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let mut visited = HashSet::new();
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let mut answer = vec![(i32::MAX, i32::MAX); n];
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answer[0] = (0, 0);
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for edge in &red_edges {
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red_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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for edge in &blue_edges {
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blue_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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while let Some((Reverse(length), node, is_red)) = nodes_heap.pop() {
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if visited.contains(&(node, is_red)) {
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continue;
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}
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visited.insert((node, is_red));
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if is_red {
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for &next in &blue_nexts[node] {
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if answer[next].1 > length + 1 {
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answer[next].1 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, false));
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}
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}
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} else {
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for &next in &red_nexts[node] {
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if answer[next].0 > length + 1 {
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answer[next].0 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, true));
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}
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}
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}
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}
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answer
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.into_iter()
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.map(|(red, blue)| {
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if red.min(blue) == i32::MAX {
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-1
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} else {
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red.min(blue)
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}
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})
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.collect()
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}
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}
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```

‎Problemset/1129-Shortest Path with Alternating Colors/Solution.rs

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use std::cmp::Reverse;
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use std::collections::BinaryHeap;
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use std::collections::HashSet;
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impl Solution {
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pub fn shortest_alternating_paths(
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n: i32,
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red_edges: Vec<Vec<i32>>,
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blue_edges: Vec<Vec<i32>>,
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) -> Vec<i32> {
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let n = n as usize;
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let mut red_nexts = vec![vec![]; n];
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let mut blue_nexts = vec![vec![]; n];
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let mut nodes_heap = BinaryHeap::from([(Reverse(0), 0, true), (Reverse(0), 0, false)]);
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let mut visited = HashSet::new();
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let mut answer = vec![(i32::MAX, i32::MAX); n];
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answer[0] = (0, 0);
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for edge in &red_edges {
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red_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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for edge in &blue_edges {
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blue_nexts[edge[0] as usize].push(edge[1] as usize);
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}
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while let Some((Reverse(length), node, is_red)) = nodes_heap.pop() {
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if visited.contains(&(node, is_red)) {
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continue;
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}
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visited.insert((node, is_red));
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if is_red {
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for &next in &blue_nexts[node] {
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if answer[next].1 > length + 1 {
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answer[next].1 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, false));
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}
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}
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} else {
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for &next in &red_nexts[node] {
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if answer[next].0 > length + 1 {
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answer[next].0 = length + 1;
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nodes_heap.push((Reverse(length + 1), next, true));
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}
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}
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}
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}
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answer
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.into_iter()
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.map(|(red, blue)| {
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if red.min(blue) == i32::MAX {
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-1
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} else {
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red.min(blue)
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}
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})
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.collect()
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}
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}

‎README.md

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652652
[1122][1122l]|[Relative Sort Array][1122] |![rs]
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[1123][1123l]|[Lowest Common Ancestor of Deepest Leaves][1123] |![py]
654654
[1128][1128l]|[Number of Equivalent Domino Pairs][1128] |![rs]
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[1129][1129l]|[Shortest Path with Alternating Colors][1129] |![rs]
655656
[1131][1131l]|[Maximum of Absolute Value Expression][1131] |![rs]
656657
[1137][1137l]|[N-th Tribonacci Number][1137] |![rs]
657658
[1138][1138l]|[Alphabet Board Path][1138] |![rb]&nbsp;&nbsp;![rs]
@@ -2004,6 +2005,7 @@
20042005
[1122]:Problemset/1122-Relative%20Sort%20Array/README.md#1122-relative-sort-array
20052006
[1123]:Problemset/1123-Lowest%20Common%20Ancestor%20of%20Deepest%20Leaves/README.md#1123-lowest-common-ancestor-of-deepest-leaves
20062007
[1128]:Problemset/1128-Number%20of%20Equivalent%20Domino%20Pairs/README.md#1128-number-of-equivalent-domino-pairs
2008+
[1129]:Problemset/1129-Shortest%20Path%20with%20Alternating%20Colors/README.md#1129-shortest-path-with-alternating-colors
20072009
[1131]:Problemset/1131-Maximum%20of%20Absolute%20Value%20Expression/README.md#1131-maximum-of-absolute-value-expression
20082010
[1137]:Problemset/1137-N-th%20Tribonacci%20Number/README.md#1137-n-th-tribonacci-number
20092011
[1138]:Problemset/1138-Alphabet%20Board%20Path/README.md#1138-alphabet-board-path
@@ -3359,6 +3361,7 @@
33593361
[1122l]:https://leetcode.com/problems/relative-sort-array/
33603362
[1123l]:https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/
33613363
[1128l]:https://leetcode.com/problems/number-of-equivalent-domino-pairs/
3364+
[1129l]:https://leetcode.com/problems/shortest-path-with-alternating-colors/
33623365
[1131l]:https://leetcode.com/problems/maximum-of-absolute-value-expression/
33633366
[1137l]:https://leetcode.com/problems/n-th-tribonacci-number/
33643367
[1138l]:https://leetcode.com/problems/alphabet-board-path/

‎README_CN.md

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@@ -652,6 +652,7 @@
652652
[1122][1122l]|[数组的相对排序][1122] |![rs]
653653
[1123][1123l]|[最深叶节点的最近公共祖先][1123] |![py]
654654
[1128][1128l]|[等价多米诺骨牌对的数量][1128] |![rs]
655+
[1129][1129l]|[颜色交替的最短路径][1129] |![rs]
655656
[1131][1131l]|[绝对值表达式的最大值][1131] |![rs]
656657
[1137][1137l]|[第 N 个泰波那契数][1137] |![rs]
657658
[1138][1138l]|[字母板上的路径][1138] |![rb]&nbsp;&nbsp;![rs]
@@ -2004,6 +2005,7 @@
20042005
[1122]:Problemset/1122-Relative%20Sort%20Array/README_CN.md#1122-数组的相对排序
20052006
[1123]:Problemset/1123-Lowest%20Common%20Ancestor%20of%20Deepest%20Leaves/README_CN.md#1123-最深叶节点的最近公共祖先
20062007
[1128]:Problemset/1128-Number%20of%20Equivalent%20Domino%20Pairs/README_CN.md#1128-等价多米诺骨牌对的数量
2008+
[1129]:Problemset/1129-Shortest%20Path%20with%20Alternating%20Colors/README_CN.md#1129-颜色交替的最短路径
20072009
[1131]:Problemset/1131-Maximum%20of%20Absolute%20Value%20Expression/README_CN.md#1131-绝对值表达式的最大值
20082010
[1137]:Problemset/1137-N-th%20Tribonacci%20Number/README_CN.md#1137-第-n-个泰波那契数
20092011
[1138]:Problemset/1138-Alphabet%20Board%20Path/README_CN.md#1138-字母板上的路径
@@ -3359,6 +3361,7 @@
33593361
[1122l]:https://leetcode.cn/problems/relative-sort-array/
33603362
[1123l]:https://leetcode.cn/problems/lowest-common-ancestor-of-deepest-leaves/
33613363
[1128l]:https://leetcode.cn/problems/number-of-equivalent-domino-pairs/
3364+
[1129l]:https://leetcode.cn/problems/shortest-path-with-alternating-colors/
33623365
[1131l]:https://leetcode.cn/problems/maximum-of-absolute-value-expression/
33633366
[1137l]:https://leetcode.cn/problems/n-th-tribonacci-number/
33643367
[1138l]:https://leetcode.cn/problems/alphabet-board-path/

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