From 3673870d3609f9947c40d631570b6715149a2d60 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Thu, 22 Feb 2024 15:55:03 +0800
Subject: [PATCH] 'commit'

---
 .../LanQiaoBei/LanQiao15STEMA20231217/11.cpp  | 38 ++++++++-----------
 .../LanQiaoBei/LanQiao15STEMA20231217/11.in   |  9 +++++
 2 files changed, 24 insertions(+), 23 deletions(-)
 create mode 100644 TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in

diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
index 58f1904..7c08cb7 100644
--- a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
@@ -1,8 +1,8 @@
 #include <bits/stdc++.h>
 using namespace std;
 const int N = 100010;
-vector<int> g[N], f[N]; // 邻接表,一个正图，一个反图
-int c[N];               // 每个树洞中松鼠的数量
+vector<int> g[N]; // 邻接表,一个正图，一个反图
+int c[N];         // 每个树洞中松鼠的数量
 /*
 测试用例
 4
@@ -15,19 +15,21 @@ int c[N];               // 每个树洞中松鼠的数量
 2 4
 2
 */
-int dfs1(int u, int k) {
-    if (k == 0) return 0;
+int dfs(int u, int fa, int k) {
+    if (k == 0) return c[u];
     int sum = c[u];
-    for (int i = 0; i < g[u].size(); i++) sum += dfs1(g[u][i], k - 1);
-    return sum;
-}
-int dfs2(int u, int k) {
-    if (k == 0) return 0;
-    int sum = c[u];
-    for (int i = 0; i < f[u].size(); i++) sum += dfs2(f[u][i], k - 1);
+    for (int i = 0; i < g[u].size(); i++) {
+        if (g[u][i] == fa) continue;
+        sum += dfs(g[u][i], u, k - 1);
+    }
     return sum;
 }
+
 int main() {
+#ifndef ONLINE_JUDGE
+    freopen("11.in", "r", stdin);
+#endif
+
     int n; // 树洞的数量
     cin >> n;
 
@@ -37,21 +39,11 @@ int main() {
         int a, b; // 表示两个树洞相连接
         cin >> a >> b;
         g[a].push_back(b); // 用邻接表保存一下某个树洞与哪两个其它树洞相连接
-        f[b].push_back(a); // 建反图
+        g[b].push_back(a);
     }
 
     int k;
     cin >> k;
-
-    for (int i = 1; i <= n; i++) {
-        // 计算i号树洞与它在k个距离内的所有树洞的松鼠个数和
-        // 1、向下k个距离 k<=20
-        int sum = dfs1(i, k);
-        // 2、向上k个距离 k<=20
-        sum += dfs2(i, k);
-
-        cout << sum - c[i] << endl;
-    }
-
+    for (int i = 1; i <= n; i++) cout << dfs(i, -1, k) << endl;
     return 0;
 }
\ No newline at end of file
diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in
new file mode 100644
index 0000000..14224a9
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.in
@@ -0,0 +1,9 @@
+4
+5
+3
+6
+1
+1 2
+1 3
+2 4
+2
\ No newline at end of file