結果

問題 No.1442 I-wate Shortest Path Problem
コンテスト
ユーザー chineristAC
提出日時 2021-03-26 22:50:19
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 4,157 bytes
コンパイル時間 445 ms
コンパイル使用メモリ 82,400 KB
実行使用メモリ 191,840 KB
最終ジャッジ日時 2024-11-29 00:49:42
合計ジャッジ時間 32,940 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge3
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
sample AC * 2
other AC * 2 WA * 20 TLE * 3
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 # 

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)

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

# Euler Tour の構築

depth = [0]*N
depth_dist = [0]*N
cnt = [0 for v in range(N)]
stack = [(0,0,0)]
S = []
F = [0]*N
while stack:
    v,d,di = stack[-1]
    if cnt[v]==0:
        F[v] = len(S)
        depth[v] = d
        depth_dist[v] = di
        S.append(v)
    if cnt[v] == len(edge[v]):
        stack.pop()
        continue
    nv,c = edge[v][cnt[v]]
    cnt[v] += 1
    stack.append((nv,d+1,di+c))


# 存在しない範囲は深さが他よりも大きくなるようにする
INF = (N, None)

# LCAを計算するクエリの前計算
M = 2*N
M0 = 2**(M-1).bit_length()
data = [INF]*(2*M0)
for i, v in enumerate(S):
    data[M0-1+i] = (depth[v], i)
for i in range(M0-2, -1, -1):
    data[i] = min(data[2*i+1], data[2*i+2])

# LCAの計算 (generatorで最小値を求める)
def _query(a, b):
    res = INF
    a += M0; b += M0
    while a < b:
        if b & 1:
            b -= 1
            res = min(res,data[b-1])
        if a & 1:
            res = min(res,data[a-1])
            a += 1
        a >>= 1; b >>= 1
    return res

# LCAの計算 (外から呼び出す関数)
def lca(u, v):
    fu = F[u]; fv = F[v]
    if fu > fv:
        fu, fv = fv, fu
    idx = _query(fu,fv+1)
    return S[idx[1]]

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