結果
問題 | No.1442 I-wate Shortest Path Problem |
ユーザー | chineristAC |
提出日時 | 2021-03-26 22:27:37 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 2,844 ms / 3,000 ms |
コード長 | 3,853 bytes |
コンパイル時間 | 540 ms |
コンパイル使用メモリ | 82,048 KB |
実行使用メモリ | 185,592 KB |
最終ジャッジ日時 | 2024-05-06 16:11:06 |
合計ジャッジ時間 | 28,617 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge4 |
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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 | 48 ms
56,576 KB |
testcase_01 | AC | 54 ms
56,704 KB |
testcase_02 | AC | 229 ms
81,276 KB |
testcase_03 | AC | 292 ms
88,700 KB |
testcase_04 | AC | 221 ms
81,192 KB |
testcase_05 | AC | 149 ms
80,000 KB |
testcase_06 | AC | 310 ms
88,436 KB |
testcase_07 | AC | 203 ms
80,528 KB |
testcase_08 | AC | 264 ms
86,772 KB |
testcase_09 | AC | 261 ms
83,404 KB |
testcase_10 | AC | 363 ms
89,672 KB |
testcase_11 | AC | 292 ms
88,048 KB |
testcase_12 | AC | 1,459 ms
163,960 KB |
testcase_13 | AC | 544 ms
144,412 KB |
testcase_14 | AC | 1,064 ms
155,532 KB |
testcase_15 | AC | 931 ms
154,036 KB |
testcase_16 | AC | 1,523 ms
163,476 KB |
testcase_17 | AC | 2,710 ms
183,352 KB |
testcase_18 | AC | 2,812 ms
184,656 KB |
testcase_19 | AC | 1,734 ms
168,428 KB |
testcase_20 | AC | 2,706 ms
184,364 KB |
testcase_21 | AC | 2,844 ms
185,592 KB |
testcase_22 | AC | 518 ms
144,812 KB |
testcase_23 | AC | 1,470 ms
163,712 KB |
testcase_24 | AC | 460 ms
150,976 KB |
testcase_25 | AC | 1,514 ms
182,968 KB |
testcase_26 | AC | 593 ms
154,332 KB |
ソースコード
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")