feat: 重新规划路线

Author: yi-ge

README.md

@@ -547,6 +547,10 @@ C++标准库提供的数据结构实在是太多了，参考[C++标准库头文
 
 ### 图
 
+- [重新规划路线](src/graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero.cpp)  [深度优先搜索, 广度优先搜索, 图]
+
+  - LeetCode 1466. 重新规划路线 <https://leetcode.cn/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero>
+
 - [最小化旅行的价格总和](src/graphs/minimize_the_total_price_of_the_trips.cpp)  [树, 深度优先搜索, 图, 数组, 动态规划]
 
   - LeetCode 2646. 最小化旅行的价格总和 <https://leetcode.cn/problems/minimize-the-total-price-of-the-trips>

images/graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero.jpeg

@@ -0,0 +1,35 @@
+// 重新规划路线
+// https://leetcode.cn/problems/reorder-routes-to-make-all-paths-lead-to-the-city-zero
+// INLINE
+// ../../images/graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero.jpeg
+
+#include <headers.hpp>
+
+class Solution {
+private:
+  void dfs(int node, vector<bool> &visited, int &res,
+           vector<vector<pair<int, bool>>> &graph) {
+    visited[node] = true;
+    for (auto &[next, is_out] : graph[node]) {
+      if (!visited[next]) {
+        if (is_out) {
+          res++;
+        }
+        dfs(next, visited, res, graph);
+      }
+    }
+  }
+
+public:
+  int minReorder(int n, vector<vector<int>> &connections) {
+    int res = 0;
+    vector<bool> visited(n, false);
+    vector<vector<pair<int, bool>>> graph(n);
+    for (auto &connection : connections) {
+      graph[connection[0]].push_back({connection[1], true});
+      graph[connection[1]].push_back({connection[0], false});
+    }
+    dfs(0, visited, res, graph);
+    return res;
+  }
+};

src/graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero.cpp

@@ -0,0 +1,26 @@
+#include <graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero.cpp>
+
+TEST(重新规划路线, minReorder) {
+  Solution solution;
+  // 示例 1：
+  // 输入：n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]
+  // 输出：3
+  // 解释：更改以红色显示的路线的方向，使每个城市都可以到达城市 0 。
+  int n = 6;
+  vector<vector<int>> connections = {{0, 1}, {1, 3}, {2, 3}, {4, 0}, {4, 5}};
+  EXPECT_EQ(solution.minReorder(n, connections), 3);
+
+  // 示例 2：
+  // 输入：n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]
+  // 输出：2
+  // 解释：更改以红色显示的路线的方向，使每个城市都可以到达城市 0 。
+  n = 5;
+  connections = {{1, 0}, {1, 2}, {3, 2}, {3, 4}};
+  EXPECT_EQ(solution.minReorder(n, connections), 2);
+  // 示例 3：
+  // 输入：n = 3, connections = [[1,0],[2,0]]
+  // 输出：0
+  n = 3;
+  connections = {{1, 0}, {2, 0}};
+  EXPECT_EQ(solution.minReorder(n, connections), 0);
+}

test/graphs/reorder_routes_to_make_all_paths_lead_to_the_city_zero_test.cpp

@@ -1,4 +1,4 @@
-// 执行编译时间：2023-12-06 09:50:10
+// 执行编译时间：2023-12-07 11:57:41
 #include <gtest/gtest.h>
 #include <lib.hpp>