#813505 (PyPy3) No.1449 新プロランド

• 問題: No.1449 新プロランド
• 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

ファイルパターン	結果
sample	AC * 3
other	AC * 26

ソースコード

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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)
 
 
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