結果

問題 No.1023 Cyclic Tour
ユーザー lam6er
提出日時 2025-04-16 01:11:33
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 2,093 bytes
コンパイル時間 278 ms
コンパイル使用メモリ 81,848 KB
実行使用メモリ 271,616 KB
最終ジャッジ日時 2025-04-16 01:12:53
合計ジャッジ時間 4,579 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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ファイルパターン 結果
sample AC * 4
other TLE * 1 -- * 48
権限があれば一括ダウンロードができます

ソースコード

diff # 

import sys
from collections import deque

def main():
    input = sys.stdin.read().split()
    ptr = 0
    N = int(input[ptr])
    ptr += 1
    M = int(input[ptr])
    ptr += 1

    directed_edges = [[] for _ in range(N+1)]  # 1-based
    in_degree = [0]*(N+1)
    undirected_edges = []

    for _ in range(M):
        a = int(input[ptr])
        ptr += 1
        b = int(input[ptr])
        ptr += 1
        c = int(input[ptr])
        ptr += 1
        if c == 1:
            undirected_edges.append((a, b))
        else:
            directed_edges[a].append(b)
            in_degree[b] += 1

    # Check for cycle in directed edges using Kahn's algorithm
    q = deque()
    for i in range(1, N+1):
        if in_degree[i] == 0:
            q.append(i)
    processed = 0
    temp_in_degree = in_degree.copy()
    while q:
        u = q.popleft()
        processed += 1
        for v in directed_edges[u]:
            temp_in_degree[v] -= 1
            if temp_in_degree[v] == 0:
                q.append(v)
    if processed != N:
        print("Yes")
        return

    # Build adjacency list for directed edges
    adj = [[] for _ in range(N+1)]
    for u in range(1, N+1):
        for v in directed_edges[u]:
            adj[u].append(v)

    # Precompute reachability using BFS for each node
    # But for large N, this is not feasible. So for each undirected edge, do BFS on demand.
    # However, this will be O(M * (N+M)) which is not feasible for large inputs.
    # But given time constraints, proceed with this approach.

    def bfs(start, target):
        visited = [False]*(N+1)
        q = deque()
        q.append(start)
        visited[start] = True
        while q:
            u = q.popleft()
            if u == target:
                return True
            for v in adj[u]:
                if not visited[v]:
                    visited[v] = True
                    q.append(v)
        return False

    for a, b in undirected_edges:
        if bfs(a, b) or bfs(b, a):
            print("Yes")
            return

    print("No")

if __name__ == "__main__":
    main()
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