結果
| 問題 |
No.1950 片道きゃっちぼーる
|
| コンテスト | |
| ユーザー |
tktk_snsn
|
| 提出日時 | 2022-05-20 22:31:43 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 1,146 ms / 3,000 ms |
| コード長 | 4,655 bytes |
| コンパイル時間 | 211 ms |
| コンパイル使用メモリ | 82,376 KB |
| 実行使用メモリ | 460,520 KB |
| 最終ジャッジ日時 | 2024-09-20 08:45:51 |
| 合計ジャッジ時間 | 12,970 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 23 |
ソースコード
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
class SCC_graph(object):
def __init__(self, n):
"""n:ノード数"""
self.n = n
self.edges = []
def add_edge(self, frm, to):
"""frm -> toへ有効辺を張る"""
self.edges.append((frm, to))
def __csr(self):
self.start = [0] * (self.n + 1)
self.elist = [0] * len(self.edges)
for frm, to in self.edges:
self.start[frm + 1] += 1
for i in range(1, self.n + 1):
self.start[i] += self.start[i - 1]
cnt = self.start[:]
for frm, to in self.edges:
self.elist[cnt[frm]] = to
cnt[frm] += 1
def __dfs(self, v):
self.low[v] = self.now_ord
self.order[v] = self.now_ord
self.now_ord += 1
self.visited.append(v)
for i in range(self.start[v], self.start[v + 1]):
to = self.elist[i]
if self.order[to] == -1:
self.__dfs(to)
self.low[v] = min(self.low[v], self.low[to])
else:
self.low[v] = min(self.low[v], self.order[to])
if self.low[v] == self.order[v]:
while self.visited:
u = self.visited.pop()
self.order[u] = self.n
self.ids[u] = self.group_num
if u == v:
break
self.group_num += 1
def _make_scc_ids(self):
self.__csr()
self.now_ord = 0
self.group_num = 0
self.visited = []
self.low = [0] * self.n
self.ids = [0] * self.n
self.order = [-1] * self.n
for i in range(self.n):
if self.order[i] == -1:
self.__dfs(i)
for i in range(self.n):
self.ids[i] = self.group_num - 1 - self.ids[i]
def scc(self):
"""
強連結成分分解O(N+M), groupsを返す
self.ids[i] -> 頂点iがトポロジカル順で何番目の成分に属するか
groups[j] -> トポロジカル順でj番目の強連結成分に属する頂点集合
"""
self._make_scc_ids()
groups = [[] for _ in range(self.group_num)]
for i in range(self.n):
groups[self.ids[i]].append(i)
return groups
def make_condensation_graph(self):
"""強連結成分間の隣接リスト、入次数/出次数のリストを返す"""
n = self.n
G = [[] for _ in range(self.group_num)]
indeg = [0] * self.group_num
outdeg = [0] * self.group_num
used = set()
for s, t in self.edges:
s = self.ids[s]
t = self.ids[t]
if s == t:
continue
if s * n + t in used:
continue
# G[s].append(t)
G[t].append(s)
indeg[t] += 1
outdeg[s] += 1
used.add(s * n + t)
return G, indeg, outdeg
class TwoSAT(SCC_graph):
def __init__(self, n):
""" n: ノード数"""
self._n = n
super().__init__(2 * n)
def add_clause(self, i, f, j, g):
""" (xi == f) or (xj == g)というクローズを追加 """
x = 2 * i + (0 if f else 1)
y = 2 * j + (1 if g else 0)
self.add_edge(x, y)
x = 2 * j + (0 if g else 1)
y = 2 * i + (1 if f else 0)
self.add_edge(x, y)
def satisfiable(self):
""" 条件を満たす割り当てが存在するか判定する """
self._make_scc_ids()
self._answer = [False] * self._n
for i in range(self._n):
if self.ids[2 * i] == self.ids[2 * i + 1]:
return False
self._answer[i] = (self.ids[2 * i] < self.ids[2 * i + 1])
return True
def answer(self):
""" 最後に読んだsatisfiableのクローズを満たす割り当てを返す """
return self._answer
N = int(input())
X = list(map(int, input().split()))
A = list(map(int, input().split()))
E = [x + a for x, a in zip(X, A)]
xtoi = {x: i for i, x in enumerate(X)}
XS = set(X)
G = SCC_graph(N)
for i in range(N):
if X[i] + A[i] in XS:
G.add_edge(i, xtoi[X[i] + A[i]])
if X[i] - A[i] in XS:
G.add_edge(i, xtoi[X[i] - A[i]])
groups = G.scc()
M = len(groups)
d = [0] * M
for i in range(M):
for j in groups[i]:
d[i] = max(d[i], E[j])
rG, _, _ = G.make_condensation_graph()
for i in reversed(range(M)):
for j in rG[i]:
d[j] = max(d[j], d[i])
ans = [0] * N
for i in range(M):
for j in groups[i]:
ans[j] = d[i] - X[j]
print(*ans, sep="\n")
tktk_snsn