From d68fcc1054b44462ccabbc33e3a6347052ac98a8 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=EC=A0=95=EC=9C=A0=ED=99=98?=
 <jeong-yuhwan@jeong-yuhwan-ui-noteubug.local>
Date: Sun, 28 Sep 2025 10:04:33 +0900
Subject: [PATCH] =?UTF-8?q?12784=EB=B2=88=20=ED=92=80=EC=9D=B4=20=EC=B6=94?=
 =?UTF-8?q?=EA=B0=80?=
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---
 Appendix E/solutions/12784.cpp | 38 ++++++++++++++++++++++++++++++----
 1 file changed, 34 insertions(+), 4 deletions(-)

diff --git a/Appendix E/solutions/12784.cpp b/Appendix E/solutions/12784.cpp
index 6c991660..dfa9a310 100644
--- a/Appendix E/solutions/12784.cpp	
+++ b/Appendix E/solutions/12784.cpp	
@@ -1,11 +1,41 @@
-// Authored by : BaaaaaaaaaaarkingDog
+// Authored by : uhwan0723
 // Co-authored by : -
-// http://boj.kr/****************
+// http://boj.kr/f9af78fd8e5246439d180420f8a57181
 #include <bits/stdc++.h>
 using namespace std;
 
+const int mn = 1005, md = 25;
+int t, n, m;
+vector<pair<int, int>> adj[mn]; // {향하는 정점, 간선의 가중치} 형태로 저장
+
+/*
+dfs(cur) = 정점 cur을 루트로 하는 서브 트리에서 cur과 리프노드들과의 연결을 끊기 위해 필요한 최소 다이너마이트의 개수.
+         = cur의 모든 자식 ch에 대해 min(cur-ch 다리 폭파 비용, dfs(cur, ch)) 값의 합
+*/
+int dfs(int par, int cur){
+  int res = 0;
+  for(auto nxt : adj[cur]){
+    if(nxt.first == par) continue;
+    res += min(nxt.second, dfs(cur, nxt.first));
+  }
+  if(cur != 1 && res == 0) res = md;
+  return res;
+}
+
 int main(void){
   ios::sync_with_stdio(0);
   cin.tie(0);
-  
-}
\ No newline at end of file
+  cin >> t;
+  while(t--){
+    cin >> n >> m;
+    while(m--){
+      int u, v, w;
+      cin >> u >> v >> w;
+      adj[u].push_back({v, w});
+      adj[v].push_back({u, w});
+    }
+    cout << dfs(-1, 1) << '\n';
+    for(int i = 1; i <= n; ++i)
+      adj[i].clear();
+  }
+}