結果
| 問題 |
No.1449 新プロランド
|
| コンテスト | |
| ユーザー |
terasa
|
| 提出日時 | 2022-11-03 16:20:55 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 1,628 ms / 2,000 ms |
| コード長 | 2,663 bytes |
| コンパイル時間 | 292 ms |
| コンパイル使用メモリ | 82,304 KB |
| 実行使用メモリ | 246,928 KB |
| 最終ジャッジ日時 | 2024-12-24 02:11:39 |
| 合計ジャッジ時間 | 19,282 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 26 |
ソースコード
from typing import List, Tuple, Optional
import sys
import itertools
import heapq
import bisect
from collections import deque, defaultdict
from functools import lru_cache, cmp_to_key
input = sys.stdin.readline
# for AtCoder Easy test
if __file__ != 'prog.py':
# sys.setrecursionlimit(10 ** 6)
pass
def readints(): return map(int, input().split())
def readlist(): return list(readints())
def readstr(): return input().rstrip()
class Dijkstra:
def __init__(self, N: int, E: List[List[Tuple[int, int]]],
start: int = 0, inf: int = 1 << 50):
self.N = N
self.E = E
self.start = start
self.inf = inf
self.C = [self.inf] * N
self.prev = [None] * N
self._calculate()
def get_cost(self, i: int) -> Optional[int]:
"""return cost to i-th vertex. return inf if the vertex is unreachable."""
return self.C[i]
def get_path(self, i) -> Optional[List[int]]:
"""return shortest path to i-th vertex if reachable otherwise None"""
if not self.reachable(i):
return None
p = []
cur = i
while cur is not None:
p.append(cur)
cur = self.prev[cur]
p.reverse()
return p
def reachable(self, i) -> bool:
"""return whether i-th vertex is reachable from start"""
return self.C[i] < self.inf
def _calculate(self) -> None:
h = [(0, self.start)]
self.C[self.start] = 0
visited = [False] * self.N
while h:
_, v = heapq.heappop(h)
if visited[v] is True:
continue
visited[v] = True
for c, d in self.E[v]:
if self.C[d] > self.C[v] + c:
self.C[d] = self.C[v] + c
self.prev[d] = v
heapq.heappush(h, (self.C[d], d))
N, M = readints()
edges = [tuple(readints()) for _ in range(M)]
T = readlist()
L = 10000
E = [[] for _ in range(N * (L + 1))]
for a, b, c in edges:
a -= 1
b -= 1
def h(i, t):
return i * (L + 1) + t
for j in range(L + 1):
if j + T[a] > L:
break
if j + T[a] == 0:
continue
cost = T[a] + c // (j + T[a])
E[h(a, j)].append((cost, h(b, j + T[a])))
for j in range(L + 1):
if j + T[b] > L:
break
if j + T[b] == 0:
continue
cost = T[b] + c // (j + T[b])
E[h(b, j)].append((cost, h(a, j + T[b])))
solver = Dijkstra(N * (L + 1), E)
ans = solver.inf
for j in range(L + 1):
ans = min(ans, solver.get_cost((N - 1) * (L + 1) + j))
print(ans)
terasa