Merge pull request #629 from LetMeFly666/3243

添加问题“3243.新增道路查询后的最短距离I”的代码和题解

Author: LetMeFly666

.vscode/settings.json

@@ -1,5 +1,8 @@
 {
     "files.associations": {
+        "*.pro": "plaintext",
+        "*.wxss": "css",
+        "*.wxml": "html",
         "array": "cpp",
         "atomic": "cpp",
         "*.tcc": "cpp",
@@ -80,7 +83,8 @@
         "iterator": "cpp",
         "memory": "cpp",
         "memory_resource": "cpp",
-        "set": "cpp"
+        "set": "cpp",
+        "math.h": "c"
     },
     "editor.mouseWheelZoom": true,
     "workbench.tree.enableStickyScroll": true,

Codes/3243-shortest-distance-after-road-addition-queries-i.cpp

@@ -0,0 +1,33 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-19 14:30:16
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-19 14:37:15
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
+        vector<int> shortest(n);
+        vector<vector<int>> fromList(n);
+        for (int i = 1; i < n; i++) {
+            fromList[i].push_back(i - 1);
+            shortest[i] = i;
+        }
+        vector<int> ans(queries.size());
+        for (int i = 0; i < queries.size(); i++) {
+            int from = queries[i][0], to = queries[i][1];
+            fromList[to].push_back(from);
+            for (int j = to; j < n; j++) {
+                for (int from : fromList[j]) {
+                    shortest[j] = min(shortest[j], shortest[from] + 1);
+                }
+            }
+            ans[i] = shortest.back();
+        }
+        return ans;
+    }
+};

Codes/3243-shortest-distance-after-road-addition-queries-i.go

@@ -0,0 +1,28 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-19 15:00:03
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-19 15:04:56
+ */
+package main
+
+func shortestDistanceAfterQueries(n int, queries [][]int) (ans []int) {
+    shortest := make([]int, n)
+    fromList := make([][]int, n)
+    for i := range shortest {
+        shortest[i] = i
+        fromList[i] = make([]int, 0)
+        fromList[i] = append(fromList[i], i - 1)
+    }
+    ans = make([]int, len(queries))
+    for i, query := range queries {
+        fromList[query[1]] = append(fromList[query[1]], query[0])
+        for j := query[1]; j < n; j++ {
+            for _, from := range fromList[j] {
+                shortest[j] = min(shortest[j], shortest[from] + 1)
+            }
+        }
+        ans[i] = shortest[n - 1]
+    }
+    return
+}

Codes/3243-shortest-distance-after-road-addition-queries-i.java

@@ -0,0 +1,32 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-11-19 14:46:11
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2024-11-19 14:58:40
+ */
+import java.util.ArrayList;
+import java.util.List;
+
+class Solution {
+    public int[] shortestDistanceAfterQueries(int n, int[][] queries) {
+        int[] shortest = new int[n];
+        List<Integer>[] fromList = new List[n];
+        for (int i = 0; i < n; i++) {
+            shortest[i] = i;
+            fromList[i] = new ArrayList<Integer>();
+            fromList[i].add(i - 1);
+        }
+        int[] ans = new int[queries.length];
+        for (int i = 0; i < queries.length; i++) {
+            int from = queries[i][0], to = queries[i][1];
+            fromList[to].add(from);
+            for (int j = to; j < n; j++) {
+                for (int from_ : fromList[j]) {
+                    shortest[j] = Math.min(shortest[j], shortest[from_] + 1);
+                }
+            }
+            ans[i] = shortest[n - 1];
+        }
+        return ans;
+    }
+}

Codes/3243-shortest-distance-after-road-addition-queries-i.py

@@ -0,0 +1,19 @@
+'''
+Author: LetMeFly
+Date: 2024-11-19 14:37:55
+LastEditors: LetMeFly.xyz
+LastEditTime: 2024-11-19 14:45:39
+'''
+from typing import List
+
+class Solution:
+    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
+        shortest = [i for i in range(n)]
+        fromList = [[i - 1] for i in range(n)]
+        ans = []
+        for from_, to in queries:
+            fromList[to].append(from_)
+            for i in range(to, n):
+                shortest[i] = min(shortest[i], min(shortest[j] + 1 for j in fromList[i]))
+            ans.append(shortest[-1])
+        return ans