Merge pull request #1155 from 0xff-dev/3108

Add solution and test-cases for problem 3108

Author: 6boris

leetcode/3101-3200/3108.Minimum-Cost-Walk-in-Weighted-Graph/1.png

@@ -1,28 +1,48 @@
 # [3108.Minimum Cost Walk in Weighted Graph][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+There is an undirected weighted graph with `n` vertices labeled from `0` to `n - 1`.
+
+You are given the integer `n` and an array `edges`, where `edges[i] = [ui, vi, wi]` indicates that there is an edge between vertices `ui` and `vi` with a weight of `wi`.
+
+A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It's important to note that a walk may visit the same edge or vertex more than once.
+
+The **cost** of a walk starting at node `u` and ending at node `v` is defined as the bitwise `AND` of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is `w0, w1, w2, ..., wk`, then the cost is calculated as `w0 & w1 & w2 & ... & wk`, where `&` denotes the bitwise `AND` operator.
+
+You are also given a 2D array `query`, where `query[i] = [si, ti]`. For each query, you need to find the minimum cost of the walk starting at vertex `si` and ending at vertex `ti`. If there exists no such walk, the answer is `-1`.
+
+Return the array `answer`, where `answer[i]` denotes the **minimum** cost of a walk for query `i`.
+
+**Example 1:**  
+
+![1](./1.png)
 
-**Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]
+
+Output: [1,-1]
+
+Explanation:
+
+To achieve the cost of 1 in the first query, we need to move on the following edges: 0->1 (weight 7), 1->2 (weight 1), 2->1 (weight 1), 1->3 (weight 7).
+
+In the second query, there is no walk between nodes 3 and 4, so the answer is -1.
 ```
 
-## 题意
-> ...
+**Example 2:**  
+
+![2](./2.png)
 
-## 题解
 
-### 思路1
-> ...
-Minimum Cost Walk in Weighted Graph
-```go
 ```
+Input: n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]
+
+Output: [0]
 
+Explanation:
+To achieve the cost of 0 in the first query, we need to move on the following edges: 1->2 (weight 1), 2->1 (weight 6), 1->2 (weight 1).
+```
 
 ## 结语
 

leetcode/3101-3200/3108.Minimum-Cost-Walk-in-Weighted-Graph/2.png

@@ -1,5 +1,50 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+type unionFind3108 struct {
+	father []int
+	weight []int
+}
+
+func (u *unionFind3108) find(x int) int {
+	if u.father[x] != x {
+		u.father[x] = u.find(u.father[x])
+	}
+	return u.father[x]
+}
+
+func (u *unionFind3108) union(x, y, w int) {
+	fx := u.find(x)
+	fy := u.find(y)
+	if fx < fy {
+		u.father[fy] = fx
+		u.weight[fx] = u.weight[fx] & u.weight[fy] & w
+	} else {
+		u.father[fx] = fy
+		u.weight[fy] = u.weight[fx] & u.weight[fy] & w
+	}
+}
+
+func Solution(n int, edges [][]int, query [][]int) []int {
+	u := &unionFind3108{father: make([]int, n), weight: make([]int, n)}
+	for i := range n {
+		u.father[i] = i
+		u.weight[i] = 0xffffffff
+	}
+
+	for _, e := range edges {
+		f, t, w := e[0], e[1], e[2]
+		u.union(f, t, w)
+	}
+	ans := make([]int, len(query))
+	for i, q := range query {
+		f, t := q[0], q[1]
+		ff := u.find(f)
+		tf := u.find(t)
+		if ff != tf {
+			ans[i] = -1
+			continue
+		}
+		ans[i] = u.weight[ff]
+	}
+	return ans
 }

leetcode/3101-3200/3108.Minimum-Cost-Walk-in-Weighted-Graph/README.md

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		n      int
+		edges  [][]int
+		query  [][]int
+		expect []int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", 5, [][]int{{0, 1, 7}, {1, 3, 7}, {1, 2, 1}}, [][]int{{0, 3}, {3, 4}}, []int{1, -1}},
+		{"TestCase2", 3, [][]int{{0, 2, 7}, {0, 1, 15}, {1, 2, 6}, {1, 2, 1}}, [][]int{{1, 2}}, []int{0}},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.n, c.edges, c.query)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v %v",
+					c.expect, got, c.n, c.edges, c.query)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }