diff --git a/test99-5-2.cpp b/test99-5-2.cpp
new file mode 100644
index 0000000..d926c33
--- /dev/null
+++ b/test99-5-2.cpp
@@ -0,0 +1,52 @@
+// lanqiao3029 卖树
+#include<bits/stdc++.h>
+using namespace std;
+using ll = long long;
+const int N = 1e5 + 9;
+vector<int> g[N];
+int dep1[N], depU[N], depV[N];
+void dfs(int x, int fa, int dep[]){
+    dep[x] = dep[fa] + 1;
+    for(const auto &y:g[x]){
+        if(y == fa) continue;
+        dfs(y, x, dep);
+    }
+}
+void solve(){
+    ll n, k, c; cin >> n >> k >> c;
+    for(int i = 1; i < n; i++){
+        int u, v; cin >> u >> v;
+        g[u].push_back(v), g[v].push_back(u);
+    }
+    dep1[0] = depU[0] = depV[0] = -1;
+    dfs(1, 0, dep1);
+    // 得到U
+    int U = 1;
+    for(int i = 1; i <= n; i++) if(dep1[i] > dep1[U]) U = i;
+    dfs(U, 0, depU);
+    int V = 1;
+    for(int i = 1; i <= n; i++) if(depU[i] > depU[V]) V = i;
+    dfs(V, 0, depV);
+    // 枚举所有点计算价值和代价
+    ll ans = 0;
+    for(int i = 1; i <= n; i++){
+        // 价值 = 以 i 为根的树的所有点的最大深度， 即max(depU[i], depV[i])
+        // 代价 = 从1走到i的代价
+        ans = max(ans, max(depU[i], depV[i])*k - dep1[i]*c);
+    }   
+    cout << ans << endl;
+
+    // 多组测试数据，需要重置
+    for(int i = 1; i <= n; i++) g[i].clear();
+}
+int main(){
+    int t; cin >> t;
+    while(t--) solve();
+    return 0;
+}
+/* test samples -> 4
+1
+3 4 5
+1 2
+1 3
+*/
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