結果

問題 No.399 動的な領主
ユーザー 山本信二山本信二
提出日時 2022-05-29 00:50:22
言語 Python3
(3.12.2 + numpy 1.26.4 + scipy 1.12.0)
結果
WA  
実行時間 -
コード長 9,097 bytes
コンパイル時間 144 ms
コンパイル使用メモリ 13,568 KB
実行使用メモリ 59,624 KB
最終ジャッジ日時 2024-09-20 23:40:03
合計ジャッジ時間 5,642 ms
ジャッジサーバーID
(参考情報)
judge3 / judge5
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 31 ms
18,592 KB
testcase_01 WA -
testcase_02 WA -
testcase_03 WA -
testcase_04 WA -
testcase_05 WA -
testcase_06 TLE -
testcase_07 -- -
testcase_08 -- -
testcase_09 -- -
testcase_10 -- -
testcase_11 -- -
testcase_12 -- -
testcase_13 -- -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
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 HL:

    #u,vを結ぶpathへのクエリはここにでも
    #  f は区間 [l,r)に対するクエリ
    def f(self,l,r):
        pass
    def merge(self,x,y):
        return x + y


    
    def __init__(self,G,root):
        self.G = G
        self.root = root
        self.N = len(G) 
        self.size = [1] * self.N     #部分木のサイズ
        self.p = [-1] * self.N        #親頂点
        self.H = [None] * self.N     #Heavy_edgeでつながる子頂点。葉ではNoneが入ってる
        self._in = [-1] * self.N     #最初に探索したときの位置
        self.out = [-1] * self.N     #部分木をでるタイミング。オイラーとはちょっと違う。
                                     #開区間  [_in[i],out[i]) がiの部分木に対応
        self.pathtop = [0] * self.N  #iの属するpathの中で最も根に近い頂点。代表にする
        self.build()
        self.build_path()
    def build(self):
        stack = [(~self.root,-1),(self.root,-1)]
        G = self.G
        size = self.size
        H = self.H
        while stack:
            now,parent = stack.pop()
            if now < 0:
                now = ~now
                _max = 0
                for v in G[now]:
                    if v == parent:continue
                    size[now] += size[v]
                    if size[v] > _max:
                        _max = size[v]
                        H[now] = v
            else:
                for v in G[now]:
                    if v == parent:continue
                    self.p[v] = now
                    stack.append((~v,now))
                    stack.append((v,now))
    def build_path(self):
        stack = [(~self.root,-1,self.root),(self.root,-1,self.root)]
        count = 0
        G = self.G
        H = self.H
        while stack:
            now,parent,top = stack.pop()
            if now >= 0:
                self._in[now] = count
                count += 1
                self.pathtop[now] = top
                h = H[now]
                if h is None:continue
                for v in G[now]:
                    if v == parent or v == h:continue
                    stack.append((~v,now,v))
                    stack.append((v,now,v))
                stack.append((~h,now,top))
                stack.append((h,now,top))
            else:
                now = ~now
                self.out[now] = count
    def lca(self,a,b):
        #最近共通先祖
        pathtop = self.pathtop
        _in = self._in
        pa = pathtop[a]
        pb = pathtop[b]
        while pa != pb:
            if _in[pa] > _in[pb]:
                a = self.p[pa]
                pa = pathtop[a]
            else:
                b = self.p[pb]
                pb = pathtop[b]
        return a if _in[a] < _in[b] else b
    def subtree_query(self,a,f = None):
        #if f is None:f = self.f
        return f(self._in[a],self.out[a])
    def subtree_array(self,a):
        return (self._in[a],self.out[a])
        #下のpath_arrayとほぼ同じ。タプルを一つだけ返す

    #f = lambda l,r:seg.flod(l,r)  とか
    #f = lambda l,r:seg.oparete_range(l,r,x) とか
    #代入して使う
    def path_query(self,a,b,f = None,merge = None):
        #if f is None:f = self.f
        #if merge is None:merge = self.merge
        pathtop = self.pathtop
        p = self.p
        _in = self._in
        pa = pathtop[a]
        pb = pathtop[b]
        ans = 0
        while pa != pb:
            if _in[pa] > _in[pb]:
                ans = merge(ans,f(_in[pa],_in[a]+1))
                a = p[pa]
                pa = pathtop[a]
            else:
                ans = merge(ans,f(_in[pb],_in[b]+1))
                b = p[pb]
                pb = pathtop[b]
        if _in[a] > _in[b]:
            a,b = b,a
        ans = merge(ans,f(_in[a],_in[b]+1))
        return ans
    def path_array(self,a,b):
        # a,b を結ぶpath、を分割した配列を返す。こっちのほうが便利かも
        #半開区間 [l,r) の集まりを返す
        #現状順番は適当
        pathtop = self.pathtop
        p = self.p
        _in = self._in
        ans = []
        pa = pathtop[a]
        pb = pathtop[b]
        while pa != pb:
            if _in[pa] > _in[pb]:
                ans.append((_in[pa],_in[a]+1))
                a = p[pa]
                pa = pathtop[a]
            else:
                ans.append((_in[pb],_in[b]+1))
                b = p[pb]
                pb = pathtop[b]
        if _in[a] > _in[b]:
            a,b = b,a
        ans.append((_in[a],_in[b]+1))
        return ans

class LazyPropSegmentTree:
    def __init__(self, lst, op, apply, comp, e, identity):
        self.n = len(lst)
        self.depth = (self.n - 1).bit_length()
        self.N = 1 << self.depth
        self.op = op # binary operation of elements
        self.apply = apply # function to apply to an element
        self.comp = comp # composition of functions
        self.e = e # identity element w.r.t. op
        self.identity = identity # identity element w.r.t. comp
        self.v, self.length = self._build(lst) # self.v is set to be 1-indexed for simplicity
        self.lazy = [self.identity] * (2 * self.N)
    
    def __getitem__(self, i):
        return self.fold(i, i+1)
    
    def _build(self, lst):
        # construction of a tree
        # total 2 * self.N elements (tree[0] is not used)
        e, N, op = self.e, self.N, self.op
        tree = [e] * N + lst + [e] * (N - self.n)
        length = [1] * (2 * N)
        for i in range(N - 1, 0, -1):
            lc, rc = i << 1, (i << 1)|1
            tree[i] = op(tree[lc], tree[rc])
            length[i] = length[lc] + length[rc]
        return tree, length
    
    def _indices(self, l, r):
        left = l + self.N; right = r + self.N
        left //= (left & (-left)); right //= (right & (-right))
        left >>= 1; right >>= 1
        while left != right:
            if left > right: yield left; left >>= 1
            else: yield right; right >>= 1
        while left > 0: yield left; left >>= 1
    
    # propagate self.lazy and self.v in a top-down manner
    def _propagate_topdown(self, *indices):
        identity, v, lazy, length, apply, comp = self.identity, self.v, self.lazy, self.length, self.apply, self.comp
        for k in reversed(indices):
            x = lazy[k]
            if x == identity: continue
            lc, rc = k << 1, (k << 1) | 1
            lazy[lc] = comp(lazy[lc], x)
            lazy[rc] = comp(lazy[rc], x)
            v[lc] = apply(v[lc], x, length[lc])
            v[rc] = apply(v[rc], x, length[rc])
            lazy[k] = identity # propagated

    # propagate self.v in a bottom-up manner
    def _propagate_bottomup(self, indices):
        v, op = self.v, self.op
        for k in indices: v[k] = op(v[k << 1], v[(k << 1)|1])

    # update for the query interval [l, r) with function x
    def update(self, l, r, x):
        *indices, = self._indices(l, r)
        self._propagate_topdown(*indices)
        
        N, v, lazy, length, apply, comp = self.N, self.v, self.lazy, self.length, self.apply, self.comp
        
        # update self.v and self.lazy for the query interval [l, r)
        left = l + N; right = r + N
        if left & 1: v[left] = apply(v[left], x, length[left]); left += 1
        if right & 1: right -= 1; v[right] = apply(v[right], x, length[right])
        left >>= 1; right >>= 1
        while left < right:
            if left & 1:
                lazy[left] = comp(lazy[left], x)
                v[left] = apply(v[left], x, length[left])
                left += 1
            if right & 1:
                right -= 1
                lazy[right] = comp(lazy[right], x)
                v[right] = apply(v[right], x, length[right])
            left >>= 1; right >>= 1
        self._propagate_bottomup(indices)
    
    # returns answer for the query interval [l, r)
    def fold(self, l, r):
        self._propagate_topdown(*self._indices(l, r))
        
        e, N, v, op = self.e, self.N, self.v, self.op
        
        # calculate the answer for the query interval [l, r)
        left = l + N; right = r + N
        L = R = e
        while left < right:
            if left & 1: # self.v[left] is the right child
                L = op(L, v[left])
                left += 1
            if right & 1: # self.v[right-1] is the left child
                right -= 1
                R = op(v[right], R)
            left >>= 1; right >>= 1
        return op(L, R)

N = int(input())
G = [[] for _ in range(N+1)]
for _ in range(N-1):
    a,b = map(int,input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)
hl = HL(G,1)
"""
print(hl._in)
print(hl.p)
print(hl.pathtop)
"""
Q = int(input())
seq = [1] * N
op = lambda x, y: x+y
apply = lambda x, f, l: x + f*l
comp = lambda f, g: f + g
e = 0
identity = 0
A = seq
lpsg = LazyPropSegmentTree(A, op, apply, comp, e, identity)
ans = []
for _ in range(Q):
    *arg, = map(int, input().split())
    s, t= arg
    lpsg.update(s-1, t, 1)
    ans.append(lpsg.fold(s-1, t))
print(sum(ans))
0