結果
| 問題 |
No.160 最短経路のうち辞書順最小
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 18:49:03 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,012 bytes |
| コンパイル時間 | 217 ms |
| コンパイル使用メモリ | 82,308 KB |
| 実行使用メモリ | 77,524 KB |
| 最終ジャッジ日時 | 2025-03-20 18:50:17 |
| 合計ジャッジ時間 | 3,260 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 8 WA * 18 |
ソースコード
import heapq
n, m, S, G = map(int, input().split())
adj = [[] for _ in range(n)]
for _ in range(m):
a, b, c = map(int, input().split())
adj[a].append((b, c))
adj[b].append((a, c))
# Dijkstra's algorithm to find shortest distances
INF = float('inf')
dist = [INF] * n
dist[S] = 0
heap = []
heapq.heappush(heap, (0, S))
while heap:
current_dist, u = heapq.heappop(heap)
if current_dist > dist[u]:
continue
for v, cost in adj[u]:
if dist[v] > dist[u] + cost:
dist[v] = dist[u] + cost
heapq.heappush(heap, (dist[v], v))
# Reconstruct the path from G to S by choosing lex smallest nodes
path = []
current = G
path.append(current)
while current != S:
min_u = None
for v, cost in adj[current]:
if dist[v] + cost == dist[current]:
if min_u is None or v < min_u:
min_u = v
current = min_u
path.append(current)
# Reverse the path to get the correct order and print
print(' '.join(map(str, path[::-1])))
lam6er