From 267d3047f04b85c50862d2da3734718c62addc4f Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Tue, 16 Jan 2024 13:07:45 +0800
Subject: [PATCH] 'commit'

---
 TangDou/Topic/HuanGenDp/POJ3585.cpp | 26 ++++++++++++++------------
 1 file changed, 14 insertions(+), 12 deletions(-)

diff --git a/TangDou/Topic/HuanGenDp/POJ3585.cpp b/TangDou/Topic/HuanGenDp/POJ3585.cpp
index 7b21b01..bc8b51d 100644
--- a/TangDou/Topic/HuanGenDp/POJ3585.cpp
+++ b/TangDou/Topic/HuanGenDp/POJ3585.cpp
@@ -13,31 +13,31 @@ void add(int a, int b, int c = 0) {
     e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
 }
 
-int d[N], f[N];
+int f[N], g[N];
 
+// 以1号点为根， 由子推父，先递归，再统计
 void dfs1(int u, int fa) {
     for (int i = h[u]; ~i; i = ne[i]) {
         int v = e[i];
         if (v == fa) continue;
         dfs1(v, u);
-        if (du[v] == 1)
-            d[u] += w[i];
+        if (du[v] == 1) // 如果v的度为1，也就是叶子
+            f[u] += w[i];
         else
-            d[u] += min(d[v], w[i]);
+            f[u] += min(f[v], w[i]);
     }
-    return;
 }
 void dfs2(int u, int fa) {
     for (int i = h[u]; ~i; i = ne[i]) {
         int v = e[i];
         if (v == fa) continue;
         if (du[u] == 1)
-            f[v] = d[v] + w[i];
+            g[v] = f[v] + w[i];
         else
-            f[v] = d[v] + min(f[u] - min(d[v], w[i]), w[i]);
+            g[v] = f[v] + min(g[u] - min(f[v], w[i]), w[i]);
+        // 先计算再递归
         dfs2(v, u);
     }
-    return;
 }
 
 int main() {
@@ -50,8 +50,8 @@ int main() {
         memset(h, -1, sizeof h);
         idx = 0;
         memset(du, 0, sizeof du);
-        memset(d, 0, sizeof d);
         memset(f, 0, sizeof f);
+        memset(g, 0, sizeof g);
 
         cin >> n;
 
@@ -61,13 +61,15 @@ int main() {
             add(a, b, c), add(b, a, c);
             du[a]++, du[b]++; // 记录入度,无向图就不谈入度和出度了
         }
-
+        // 第一遍dfs
         dfs1(1, 0);
-        f[1] = d[1];
+        g[1] = f[1];
+
+        // 第二遍dfs
         dfs2(1, 0);
 
         int ans = 0;
-        for (int i = 1; i <= n; i++) ans = max(ans, f[i]);
+        for (int i = 1; i <= n; i++) ans = max(ans, g[i]);
         cout << ans << endl;
     }
     return 0;