結果

問題 No.1442 I-wate Shortest Path Problem
ユーザー chineristACchineristAC
提出日時 2021-03-26 22:27:37
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 2,678 ms / 3,000 ms
コード長 3,853 bytes
コンパイル時間 263 ms
コンパイル使用メモリ 86,900 KB
実行使用メモリ 189,088 KB
最終ジャッジ日時 2023-08-19 08:59:34
合計ジャッジ時間 28,271 ms
ジャッジサーバーID
(参考情報)
judge11 / judge15
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 99 ms
73,776 KB
testcase_01 AC 97 ms
74,196 KB
testcase_02 AC 334 ms
86,044 KB
testcase_03 AC 345 ms
92,820 KB
testcase_04 AC 257 ms
86,284 KB
testcase_05 AC 188 ms
84,392 KB
testcase_06 AC 348 ms
93,992 KB
testcase_07 AC 252 ms
85,284 KB
testcase_08 AC 297 ms
92,332 KB
testcase_09 AC 297 ms
88,620 KB
testcase_10 AC 405 ms
93,952 KB
testcase_11 AC 345 ms
94,424 KB
testcase_12 AC 1,500 ms
169,936 KB
testcase_13 AC 587 ms
149,444 KB
testcase_14 AC 1,076 ms
159,588 KB
testcase_15 AC 961 ms
159,116 KB
testcase_16 AC 1,507 ms
168,420 KB
testcase_17 AC 2,567 ms
188,396 KB
testcase_18 AC 2,575 ms
188,396 KB
testcase_19 AC 1,603 ms
173,756 KB
testcase_20 AC 2,489 ms
188,600 KB
testcase_21 AC 2,678 ms
189,088 KB
testcase_22 AC 492 ms
148,536 KB
testcase_23 AC 1,400 ms
167,904 KB
testcase_24 AC 466 ms
154,208 KB
testcase_25 AC 1,427 ms
187,408 KB
testcase_26 AC 608 ms
158,420 KB
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ソースコード

diff #

class Dijkstra():
    class Edge():
        def __init__(self, _to, _cost):
            self.to = _to
            self.cost = _cost

    def __init__(self, V):
        self.G = [[] for i in range(V)]
        self._E = 0
        self._V = V

    @property
    def E(self):
        return self._E

    @property
    def V(self):
        return self._V

    def add_edge(self, _from, _to, _cost):
        self.G[_from].append(self.Edge(_to, _cost))
        self._E += 1

    def shortest_path(self, start):
        import heapq
        que = []
        d = [10**15] * self.V
        if type(start)==int:
            s = start
            d[s] = 0
            heapq.heappush(que, (0, s))
        else:
            for s in start:
                d[s] = 0
                heapq.heappush(que,(0,s))

        while len(que) != 0:
            cost, v = heapq.heappop(que)
            if d[v] < cost: continue

            for i in range(len(self.G[v])):
                e = self.G[v][i]
                if d[e.to] > d[v] + e.cost:
                    d[e.to] = d[v] + e.cost
                    heapq.heappush(que, (d[e.to], e.to))
        return d

import sys,random,bisect
from collections import deque,defaultdict
from heapq import heapify,heappop,heappush
from itertools import permutations
from math import gcd

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

N,K = mi()
edge = [[] for i in range(N)]
tree = Dijkstra(N)
for _ in range(N-1):
    a,b,c = mi()
    edge[a-1].append((b-1,c))
    edge[b-1].append((a-1,c))
    tree.add_edge(a-1,b-1,c)
    tree.add_edge(b-1,a-1,c)

air = []
air_city = []
for _ in range(K):
    m,p = mi()
    air.append((m,p))
    air_city.append([int(a)-1 for a in input().split()])

dist_from_air = [[10**17 for i in range(N)] for a in range(K)]
for a in range(K):
    dist_from_air[a] = tree.shortest_path(air_city[a])

airs = Dijkstra(K)
for i in range(K):
    for j in range(K):
        tmp = 10**17
        for _from in air_city[i]:
            tmp = min(dist_from_air[j][_from],tmp)
        tmp += air[j][1]
        airs.add_edge(i,j,tmp)

dist = [airs.shortest_path(i) for i in range(K)]


# N: 頂点数
# G[v]: 頂点vの子頂点 (親頂点は含まない)
#
# - construct
# prv[u] = v: 頂点uの一つ上の祖先頂点v
# - lca
# kprv[k][u] = v: 頂点uの2^k個上の祖先頂点v
# depth[u]: 頂点uの深さ (根頂点は0)

prv = [-1 for i in range(N)]
deq = deque([0])
depth = [0 for i in range(N)]
depth_dist = [0 for i in range(N)]
while deq:
    v = deq.popleft()
    for nv,c in edge[v]:
        if prv[nv]==-1 and nv!=0:
            prv[nv] = v
            depth_dist[nv] = depth_dist[v] + c
            depth[nv] = depth[v] + 1
            deq.append(nv)

LV = (N-1).bit_length()
kprv = [prv]
S = prv
for k in range(LV):
    T = [0]*N
    for i in range(N):
        if S[i] is None:
            continue
        T[i] = S[S[i]]
    kprv.append(T)
    S = T

def lca(u, v):
    dd = depth[v] - depth[u]
    if dd < 0:
        u, v = v, u
        dd = -dd

    # assert depth[u] <= depth[v]
    for k in range(LV+1):
        if dd & 1:
            v = kprv[k][v]
        dd >>= 1

    # assert depth[u] == depth[v]
    if u == v:
        return u

    for k in range(LV-1, -1, -1):
        pu = kprv[k][u]; pv = kprv[k][v]
        if pu != pv:
            u = pu; v = pv

    # assert kprv[0][u] == kprv[0][v]
    return kprv[0][u]

def dist_in_tree(u,v):
    w = lca(u,v)
    return depth_dist[u] + depth_dist[v] - 2 * depth_dist[w]


ans = []
for _ in range(int(input())):
    u,v = mi()
    u,v = u-1,v-1
    res = dist_in_tree(u,v)
    for i in range(K):
        for j in range(K):
            tmp = dist_from_air[i][u] + air[i][1] + dist[i][j] + dist_from_air[j][v]
            res = min(res,tmp)
    ans.append(res)

print(*ans,sep="\n")
0