Add solution and test-cases for problem 3243

leetcode/3201-3300/3243.Shortest-Distance-After-Road-Addition-Queries-I/README.md

@@ -1,28 +1,35 @@
 # [3243.Shortest Distance After Road Addition Queries I][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
+You are given an integer `n` and a 2D integer array `queries`.
+
+There are `n` cities numbered from `0` to `n - 1`. Initially, there is a **unidirectional** road from city `i` to city `i + 1` for all `0 <= i < n - 1`.
+
+`queries[i] = [ui, vi]` represents the addition of a new **unidirectional** road from city `ui` to city `vi`. After each query, you need to find the **length** of the **shortest path** from city `0` to city `n - 1`.
 
-**Example 1:**
+Return an array `answer` where for each `i` in the range `[0, queries.length - 1]`, `answer[i]` is the length of the shortest path from city `0` to city `n - 1` after processing the **first** `i + 1` queries.
+
+**Example 1:**  
+
+![1](./1.jpg)
+![2](./2.jpg)
+![3](./3.jpg)
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: n = 5, queries = [[2,4],[0,2],[0,4]]
+Output: [3,2,1]
 ```
 
-## 题意
-> ...
+**Example 2:**  
 
-## 题解
+![4](./4.jpg)
+![5](./5.jpg)
 
-### 思路1
-> ...
-Shortest Distance After Road Addition Queries I
-```go
 ```
+Input: n = 4, queries = [[0,3],[0,2]]
 
+Output: [1,1]
+```
 
 ## 结语
 

leetcode/3201-3300/3243.Shortest-Distance-After-Road-Addition-Queries-I/Solution.go

@@ -1,5 +1,43 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(n int, queries [][]int) []int {
+	ans := make([]int, len(queries))
+	adj := make(map[int]map[int]struct{})
+	for i := range n {
+		adj[i] = make(map[int]struct{})
+	}
+	//  初始化路径
+	for i := range n - 1 {
+		adj[i][i+1] = struct{}{}
+	}
+	var bfs func() int
+	bfs = func() int {
+		queue := []int{0}
+		step := 0
+		used := map[int]struct{}{
+			0: {},
+		}
+		for len(queue) > 0 {
+			nq := make([]int, 0)
+			for _, q := range queue {
+				if q == n-1 {
+					return step
+				}
+				for rel := range adj[q] {
+					if _, ok := used[rel]; !ok {
+						nq = append(nq, rel)
+						used[rel] = struct{}{}
+					}
+				}
+			}
+			queue = nq
+			step++
+		}
+		return -1
+	}
+	for i, q := range queries {
+		adj[q[0]][q[1]] = struct{}{}
+		ans[i] = bfs()
+	}
+	return ans
 }

leetcode/3201-3300/3243.Shortest-Distance-After-Road-Addition-Queries-I/Solution_test.go

@@ -9,31 +9,31 @@ import (
 func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
-		name   string
-		inputs bool
-		expect bool
+		name    string
+		n       int
+		queries [][]int
+		expect  []int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", 5, [][]int{{2, 4}, {0, 2}, {0, 4}}, []int{3, 2, 1}},
+		{"TestCase2", 4, [][]int{{0, 3}, {0, 2}}, []int{1, 1}},
 	}
 
 	//	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.queries)
 			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",
+					c.expect, got, c.n, c.queries)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }