From 3119975208e679e1719d08ce71503e40a397365c Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=EC=A0=95=EC=9C=A0=ED=99=98?= <uhwan0723@gmail.com>
Date: Sat, 11 Oct 2025 11:16:22 +0900
Subject: [PATCH] =?UTF-8?q?22870=EB=B2=88=20=ED=92=80=EC=9D=B4=20=EC=B6=94?=
 =?UTF-8?q?=EA=B0=80?=
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---
 0x1D/solutions/22870.cpp | 61 +++++++++++++++++++++++++++++++++++++---
 1 file changed, 57 insertions(+), 4 deletions(-)

diff --git a/0x1D/solutions/22870.cpp b/0x1D/solutions/22870.cpp
index 6c991660..a5068a32 100644
--- a/0x1D/solutions/22870.cpp
+++ b/0x1D/solutions/22870.cpp
@@ -1,11 +1,64 @@
-// Authored by : BaaaaaaaaaaarkingDog
+// Authored by : uhwan0723
 // Co-authored by : -
-// http://boj.kr/****************
+// http://boj.kr/781f212c4748490487a3071dad460e4f
 #include <bits/stdc++.h>
 using namespace std;
 
+const int mn = 2e5+5, md = 2e9;
+int n, m, s, e, d[mn];
+bool pass[mn];
+vector<pair<int, int>> adj[mn];
+priority_queue<pair<int, int>> pq;
+
+int dijk(int st, int en){
+  fill(d, d+n+1, md);
+  d[st] = 0;
+  pq.push({0, st});
+  while(!pq.empty()){
+    auto cur = pq.top(); pq.pop();
+    int w = -cur.first, u = cur.second;
+    if(w != d[u]) continue;
+    for(auto nxt : adj[u]){
+      int v = nxt.first, nw = nxt.second;
+      if(d[v] <= nw+w || pass[v]) continue;
+      d[v] = nw+w;
+      pq.push({-d[v], v});
+    }
+  }
+  return d[en];
+}
+
 int main(void){
   ios::sync_with_stdio(0);
   cin.tie(0);
-  
-}
\ No newline at end of file
+  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});
+  }
+  cin >> s >> e;
+  int d1 = dijk(e, s);
+  int cur = s, dist = 0;
+  while(cur != e){
+    sort(adj[cur].begin(), adj[cur].end());
+    for(auto nxt : adj[cur]){
+      int v = nxt.first, w = nxt.second;
+      if(dist + w + d[v] != d1) continue;
+      dist += w;
+      cur = v;
+      if(cur != e) pass[cur] = true;
+      break;
+    }
+  }
+  int d2 = dijk(e, s);
+  cout << d1+d2;
+}
+
+/*
+첫 번째 다익스트라를 e에서 s로 돌리면, 각 정점들에서 e까지의 거리가 d에 저장됩니다.
+s에서부터 시작하는 cur은 사전순으로 가장 앞서는 경로에 있는 다음 정점을 찾기 위해,
+s에서 cur까지의 거리 + cur에서 nxt로의 가중치 + nxt에서 e까지의 거리 == 최단거리
+를 만족하는 가장 낮은 번호의 정점을 선택합니다.
+*/