From 6822f7dd8de367f8e4390852ce2e009a73c12e1f 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 15:32:50 +0800
Subject: [PATCH] 'commit'

---
 .../LanQiaoBei/LanQiao15STEMA20231217/11.cpp  | 57 +++++++++++++++++++
 1 file changed, 57 insertions(+)
 create mode 100644 TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp

diff --git a/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
new file mode 100644
index 0000000..58f1904
--- /dev/null
+++ b/TangDou/LanQiaoBei/LanQiao15STEMA20231217/11.cpp
@@ -0,0 +1,57 @@
+#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
+/*
+测试用例
+4
+5
+3
+6
+1
+1 2
+1 3
+2 4
+2
+*/
+int dfs1(int u, int k) {
+    if (k == 0) return 0;
+    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);
+    return sum;
+}
+int main() {
+    int n; // 树洞的数量
+    cin >> n;
+
+    for (int i = 1; i <= n; i++) cin >> c[i]; // 每个树洞中松鼠的数量
+
+    for (int i = 1; i < n; i++) {
+        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
+    }
+
+    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;
+    }
+
+    return 0;
+}
\ No newline at end of file