結果

問題 No.1744 Selfish Spies 1 (à la Princess' Perfectionism)
ユーザー lam6er
提出日時 2025-03-31 17:35:50
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 4,455 bytes
コンパイル時間 187 ms
コンパイル使用メモリ 82,288 KB
実行使用メモリ 105,180 KB
最終ジャッジ日時 2025-03-31 17:36:39
合計ジャッジ時間 5,957 ms
ジャッジサーバーID
(参考情報)
judge1 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
other AC * 21 WA * 18
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 #
プレゼンテーションモードにする

import sys
from collections import deque
def hopcroft_karp(U_size, V_size, adj):
pair_U = [0] * (U_size + 1) # Pair for spies (1-based)
pair_V = [0] * (V_size + 1) # Pair for tasks (1-based)
dist = [0] * (U_size + 1) # Distance for BFS layers
def bfs():
queue = deque()
for u in range(1, U_size + 1):
if pair_U[u] == 0:
dist[u] = 0
queue.append(u)
else:
dist[u] = float('inf')
dist[0] = float('inf') # Dummy node
while queue:
u = queue.popleft()
if dist[u] < dist[0]:
for v in adj[u]:
if dist[pair_V[v]] == float('inf'):
dist[pair_V[v]] = dist[u] + 1
queue.append(pair_V[v])
return dist[0] != float('inf')
def dfs(u):
if u != 0:
for v in adj[u]:
if dist[pair_V[v]] == dist[u] + 1:
if dfs(pair_V[v]):
pair_U[u] = v
pair_V[v] = u
return True
dist[u] = float('inf')
return False
return True
result = 0
while bfs():
for u in range(1, U_size + 1):
if pair_U[u] == 0:
if dfs(u):
result += 1
return pair_U, pair_V, result
def kosaraju_scc(graph, num_nodes):
visited = [False] * (num_nodes + 1)
order = []
# First pass to get finishing order
for node in range(1, num_nodes + 1):
if not visited[node]:
stack = [(node, False)]
while stack:
v, processed = stack.pop()
if processed:
order.append(v)
continue
if visited[v]:
continue
visited[v] = True
stack.append((v, True))
for neighbor in reversed(graph[v]):
if not visited[neighbor]:
stack.append((neighbor, False))
# Build reversed graph
reversed_graph = [[] for _ in range(num_nodes + 1)]
for u in range(1, num_nodes + 1):
for v in graph[u]:
reversed_graph[v].append(u)
# Second pass to get SCCs
visited = [False] * (num_nodes + 1)
scc_ids = [0] * (num_nodes + 1)
component_id = 0
while order:
node = order.pop()
if not visited[node]:
stack = [node]
visited[node] = True
component = []
while stack:
v = stack.pop()
component.append(v)
for neighbor in reversed_graph[v]:
if not visited[neighbor]:
visited[neighbor] = True
stack.append(neighbor)
for v in component:
scc_ids[v] = component_id
component_id += 1
return scc_ids
def main():
input = sys.stdin.read().split()
ptr = 0
N = int(input[ptr]); ptr +=1
M = int(input[ptr]); ptr +=1
L = int(input[ptr]); ptr +=1
adj = [[] for _ in range(N+1)] # adj[u] contains tasks (v) that u can do
edges = []
for _ in range(L):
S = int(input[ptr]); ptr +=1
T = int(input[ptr]); ptr +=1
adj[S].append(T)
edges.append((S, T))
# Compute maximum matching
pair_U, pair_V, max_matching = hopcroft_karp(N, M, adj)
# Build residual graph
total_nodes = N + M
residual_graph = [[] for _ in range(total_nodes +1)] # 1-based
for S, T in edges:
# For each edge (S, T)
# Check if it's matched
if pair_U[S] == T:
# Matched edge: task T → spy S in residual
residual_graph[N + T].append(S)
else:
# Unmatched edge: spy S → task T in residual
residual_graph[S].append(N + T)
# Compute SCCs
scc_ids = kosaraju_scc(residual_graph, total_nodes)
# Process each edge
for S, T in edges:
if pair_U[S] == T:
# Check if S (spy node) and T's node (N + T) are in the same SCC
spy_node = S
task_node = N + T
if scc_ids[spy_node] == scc_ids[task_node]:
print("Yes")
else:
print("No")
else:
# Unused edge, answer is Yes
print("Yes")
if __name__ == "__main__":
main()
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0