From d3a0808ed2d9c19d9697cd2e23c52629e6cd063a Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=EC=A7=80=ED=98=84=EC=84=9C?=
 <113419425+3veryDay@users.noreply.github.com>
Date: Tue, 13 Jan 2026 08:43:11 +0900
Subject: [PATCH] =?UTF-8?q?260113=20:=20[BOJ=2020183]=20=EA=B3=A8=EB=AA=A9?=
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---
 _hyunseo/20183.py | 60 +++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 60 insertions(+)
 create mode 100644 _hyunseo/20183.py

diff --git a/_hyunseo/20183.py b/_hyunseo/20183.py
new file mode 100644
index 00000000..e58f3f8d
--- /dev/null
+++ b/_hyunseo/20183.py
@@ -0,0 +1,60 @@
+#260113 : [BOJ 20183] 골목 대장 호석 - 효율성 2
+
+import sys, heapq
+input = sys.stdin.readline
+
+N, M, A, B, C = map(int, input().split())
+
+
+adj = [[] for _ in range(N+1)]
+MAX_COST = 0
+for _ in range(M ) :
+    a, b, cost = map(int, input().split())
+    MAX_COST = max(cost, MAX_COST)
+    adj[b].append((a, cost))
+    adj[a].append((b, cost))
+
+def solve(limit) :
+    # 주어진 limit이 내가 견딜 수 있는 최대 수치심일 때
+    # 내가 가지고 있는 돈 C로 이 길 돌파가 가능한지
+    # 가능하면 true, 불가능하면 false
+
+    # 다익스트라 초기화
+    distances = [float('inf')] * (N + 1)
+    distances[A] = 0
+    pq = [(0, A)]  # (누적 비용, 현재 노드)
+
+    while pq:
+        curr_dist, curr_node = heapq.heappop(pq)
+
+        if distances[curr_node] < curr_dist:
+            continue
+        
+        if curr_node == B: # 목적지 도착 시 조기 종료 가능
+            break
+
+        for next_node, weight in adj[curr_node]:
+            # 이번 경로의 가중치(수치심)가 limit보다 크면 아예 지나가지 않음
+            if weight <= limit:
+                new_dist = curr_dist + weight
+                if new_dist < distances[next_node]:
+                    distances[next_node] = new_dist
+                    heapq.heappush(pq, (new_dist, next_node))
+    
+    # 도착점까지의 최단 거리 합이 예산 C 이내?
+    return distances[B] <= C
+
+
+
+l, r = 0, MAX_COST
+ans = -1 
+while l <= r :
+    mid = (l + r ) // 2
+
+    if solve(mid) :
+        ans = mid
+        r = mid -1
+    else :
+        l = mid + 1
+
+print(ans)