結果
| 問題 |
No.1023 Cyclic Tour
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 13:47:26 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 3,171 bytes |
| コンパイル時間 | 173 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 349,512 KB |
| 最終ジャッジ日時 | 2025-06-12 13:48:00 |
| 合計ジャッジ時間 | 20,021 ms |
|
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 4 |
| other | AC * 18 WA * 4 TLE * 2 -- * 25 |
ソースコード
import sys
from sys import stdin
from collections import defaultdict, deque
def main():
sys.setrecursionlimit(1 << 25)
n, m = map(int, stdin.readline().split())
directed_edges = []
undirected_edges = []
for _ in range(m):
a, b, c = map(int, stdin.readline().split())
if c == 1:
undirected_edges.append((a, b))
else:
directed_edges.append((a, b))
# Step 1: Check if directed graph has a cycle
adj = [[] for _ in range(n+1)]
for a, b in directed_edges:
adj[a].append(b)
index = 0
indices = [0] * (n + 1)
low = [0] * (n + 1)
on_stack = [False] * (n + 1)
stack = []
sccs = []
has_cycle = False
def strongconnect(v):
nonlocal index
index += 1
indices[v] = index
low[v] = index
stack.append(v)
on_stack[v] = True
for w in adj[v]:
if indices[w] == 0:
strongconnect(w)
low[v] = min(low[v], low[w])
elif on_stack[w]:
low[v] = min(low[v], indices[w])
if low[v] == indices[v]:
scc = []
while True:
w = stack.pop()
on_stack[w] = False
scc.append(w)
if w == v:
break
sccs.append(scc)
if len(scc) >= 2:
nonlocal has_cycle
has_cycle = True
for v in range(1, n + 1):
if indices[v] == 0:
strongconnect(v)
if has_cycle:
print("Yes")
return
# Step 2: Check if any undirected edge's endpoints are in the same SCC of directed graph
node_to_scc = {}
for i, scc in enumerate(sccs):
for node in scc:
node_to_scc[node] = i
for a, b in undirected_edges:
if node_to_scc.get(a, -1) == node_to_scc.get(b, -2):
print("Yes")
return
# Step 3: Check if undirected edges form a cycle
parent = list(range(n + 1))
def find(u):
while parent[u] != u:
parent[u] = parent[parent[u]]
u = parent[u]
return u
for a, b in undirected_edges:
u_root = find(a)
v_root = find(b)
if u_root == v_root:
print("Yes")
return
parent[v_root] = u_root
# Step 4: Check for each undirected edge if u can reach v or v can reach u in directed graph
adj_dir = [[] for _ in range(n + 1)]
for a, b in directed_edges:
adj_dir[a].append(b)
# Precompute for each node, the nodes it can reach using BFS
reachable = [set() for _ in range(n + 1)]
for u in range(1, n + 1):
visited = set()
q = deque([u])
visited.add(u)
while q:
current = q.popleft()
for v in adj_dir[current]:
if v not in visited:
visited.add(v)
q.append(v)
reachable[u] = visited
for a, b in undirected_edges:
if b in reachable[a] or a in reachable[b]:
print("Yes")
return
print("No")
if __name__ == "__main__":
main()
gew1fw