Submission #7490368


Source Code Expand

import sys
sys.setrecursionlimit(10**8)
def resolve():
    class LCA(object):
        def __init__(self,edges,root=0): # edges : adjacence list
            self.edges=edges
            self.root=root
            self.n=len(edges) # n=|V|
            self.logn=(self.n-1).bit_length() # logn = ceil(log2(n))
            # initialization
            self.depth=[float("inf") if i!=root else 0 for i in range(self.n)]
            self.parent=[[-1]*self.logn for _ in range(self.n)] # parent[u][i]=v : uの2^i世代上がv(なければ-1)
            # construct
            self.__dfs(-1,root,0)
            self.__doubling()

        def __dfs(self,par,cur,dep):
            self.depth[cur]=dep
            self.parent[cur][0]=par
            for v in self.edges[cur]:
                if v!=par:
                    self.__dfs(cur,v,dep+1)

        def __doubling(self):
            for i in range(1,self.logn):
                for v in range(self.n):
                    if self.parent[v][i-1]!=-1: # 半分遡った時点で親が存在しなければ-1のままにしておく。-1が返るとlistの末尾の値の取得になってしまう
                        self.parent[v][i]=self.parent[self.parent[v][i-1]][i-1]

        def get(self,u,v): # uとvとのLCAを返す
            dd=self.depth[v]-self.depth[u]
            if dd<0: # vの方が深いようにする
                u,v=v,u
                dd=-dd

            for i in range(self.logn): # dd分だけvを遡らせる
                if dd&1: # 各bitで判定
                    v=self.parent[v][i]
                dd>>=1

            if u==v: return u # 高さ揃えた時点で一致してたら終わり

            for i in range(self.logn-1,-1,-1): # そうでなければ上から二分探索
                pu,pv=self.parent[u][i],self.parent[v][i]
                if pu!=pv:
                    u,v=pu,pv
            return self.parent[u][0]

    #%%
    # given data
    N=int(input())
    E=[[] for _ in range(N)]
    root=0
    for _ in range(N-1):
        u,v=map(lambda x:int(x)-1,input().split())
        E[u].append(v)
        E[v].append(u)
    lca=LCA(E)
    Q=int(input())
    for _ in range(Q):
        u,v=map(lambda x:int(x)-1,input().split())
        print(lca.depth[u]+lca.depth[v]-2*lca.depth[lca.get(u,v)]+1)
resolve()

Submission Info

Submission Time
Task D - 閉路
User moni0627
Language PyPy3 (2.4.0)
Score 30
Code Size 2384 Byte
Status TLE
Exec Time 2112 ms
Memory 168736 KB

Judge Result

Set Name Sample Subtask1 Subtask2
Score / Max Score 0 / 0 30 / 30 0 / 70
Status
AC × 3
AC × 12
AC × 21
TLE × 6
Set Name Test Cases
Sample subtask0_sample01.txt, subtask0_sample02.txt, subtask0_sample03.txt
Subtask1 subtask1_01.txt, subtask1_02.txt, subtask1_03.txt, subtask1_04.txt, subtask1_05.txt, subtask1_06.txt, subtask1_07.txt, subtask1_08.txt, subtask1_09.txt, subtask1_10.txt, subtask1_11.txt, subtask1_12.txt
Subtask2 subtask0_sample01.txt, subtask0_sample02.txt, subtask0_sample03.txt, subtask1_01.txt, subtask1_02.txt, subtask1_03.txt, subtask1_04.txt, subtask1_05.txt, subtask1_06.txt, subtask1_07.txt, subtask1_08.txt, subtask1_09.txt, subtask1_10.txt, subtask1_11.txt, subtask1_12.txt, subtask2_01.txt, subtask2_02.txt, subtask2_03.txt, subtask2_04.txt, subtask2_05.txt, subtask2_06.txt, subtask2_07.txt, subtask2_08.txt, subtask2_09.txt, subtask2_10.txt, subtask2_11.txt, subtask2_12.txt
Case Name Status Exec Time Memory
subtask0_sample01.txt AC 178 ms 38512 KB
subtask0_sample02.txt AC 173 ms 38256 KB
subtask0_sample03.txt AC 172 ms 38256 KB
subtask1_01.txt AC 1042 ms 168736 KB
subtask1_02.txt AC 1042 ms 165152 KB
subtask1_03.txt AC 184 ms 38256 KB
subtask1_04.txt AC 183 ms 38256 KB
subtask1_05.txt AC 232 ms 41456 KB
subtask1_06.txt AC 217 ms 40304 KB
subtask1_07.txt AC 925 ms 89120 KB
subtask1_08.txt AC 987 ms 94240 KB
subtask1_09.txt AC 1062 ms 101664 KB
subtask1_10.txt AC 1104 ms 105632 KB
subtask1_11.txt AC 1053 ms 106656 KB
subtask1_12.txt AC 1022 ms 103328 KB
subtask2_01.txt AC 1979 ms 167328 KB
subtask2_02.txt AC 1984 ms 167328 KB
subtask2_03.txt AC 1249 ms 56280 KB
subtask2_04.txt AC 1277 ms 57432 KB
subtask2_05.txt AC 1399 ms 60376 KB
subtask2_06.txt AC 1364 ms 60124 KB
subtask2_07.txt TLE 2111 ms 107040 KB
subtask2_08.txt TLE 2111 ms 107168 KB
subtask2_09.txt TLE 2112 ms 110240 KB
subtask2_10.txt TLE 2112 ms 111776 KB
subtask2_11.txt TLE 2112 ms 113056 KB
subtask2_12.txt TLE 2112 ms 117024 KB