codeforces：E2. Array and Segments (Hard version)【线段树 + 区间修改】

分析

思路很简单
遍历每个作为最大值，然后区间不包含当前最大值的都可以减掉
easy version就可以这样暴力解决
然后求出最大差值

暴力解法

import sys
input = sys.stdin.readlinen, m = list(map(int, input().split()))
a = list(map(int, input().split()))
intervals = []
for _ in range(m):l, r = n, m = list(map(int, input().split()))intervals.append([l, r])if m == 0:print(max(a) - min(a))print(0)print()
else:# fix maxmaxdiff = -0xffffffffchooseNum = 0chooseIdxs = []for maxIdx, maxn in enumerate(a):tmpNum = 0tmpIdxs = []tmp_a = a[:]cnt = 0for l, r in intervals:cnt += 1if l - 1 <= maxIdx <= r - 1:continuefor i in range(l, r + 1):tmp_a[i - 1] -= 1tmpNum += 1tmpIdxs.append(cnt)if maxn - min(tmp_a) > maxdiff:maxdiff = maxn - min(tmp_a)chooseNum = tmpNumchooseIdxs = tmpIdxsprint(maxdiff)print(chooseNum)print(*chooseIdxs)

优化

如果n和m变大
这个方法就失效了
我们要用线段树进行区间修改
特别地，我们考虑全部都选择
然后遍历maxn节点（从左到右）
以它i为左端点的所有区间都可以加上（不选择）
然后所以i - 1（当前为i）为右区间的都可以删掉
对于那种横跨i的即l <= i = r的不用删
这样的算法有一个好处就是可以保证当前包含i的所有区间都没有选择
然后线段树查询即可

segTree solution

import sys
from functools import reduce
from collections import defaultdict
input = sys.stdin.readlineclass SegTree:'''支持增量更新，覆盖更新，序列更新，任意RMQ操作基于二叉树实现初始化：O(1)增量更新或覆盖更新的单次操作复杂度：O(log k)序列更新的单次复杂度：O(n)'''def __init__(self, f1, f2, l, r, v=0):'''初始化线段树[left,right)f1,f2示例：线段和:f1=lambda a,b:a+bf2=lambda a,n:a*n线段最大值:f1=lambda a,b:max(a,b)f2=lambda a,n:a线段最小值:f1=lambda a,b:min(a,b)f2=lambda a,n:a'''self.ans = f2(v, r - l)self.f1 = f1self.f2 = f2self.l = l  # leftself.r = r  # rightself.v = v  # init valueself.lazy_tag = 0  # Lazy tagself.left = None  # SubTree(left,bottom)self.right = None  # SubTree(right,bottom)@propertydef mid_h(self):return (self.l + self.r) // 2def create_subtrees(self):midh = self.mid_hif not self.left and midh > self.l:self.left = SegTree(self.f1, self.f2, self.l, midh)if not self.right:self.right = SegTree(self.f1, self.f2, midh, self.r)def init_seg(self, M):'''将线段树的值初始化为矩阵Matrx输入保证Matrx与线段大小一致'''m0 = M[0]self.lazy_tag = 0for a in M:if a != m0:breakelse:self.v = m0self.ans = self.f2(m0, len(M))return self.ansself.v = '#'midh = self.mid_hself.create_subtrees()self.ans = self.f1(self.left.init_seg(M[:midh - self.l]), self.right.init_seg(M[midh - self.l:]))return self.ansdef cover_seg(self, l, r, v):'''将线段[left,right)覆盖为val'''if self.v == v or l >= self.r or r <= self.l:return self.ansif l <= self.l and r >= self.r:self.v = vself.lazy_tag = 0self.ans = self.f2(v, self.r - self.l)return self.ansself.create_subtrees()if self.v != '#':if self.left:self.left.v = self.vself.left.ans = self.f2(self.v, self.left.r - self.left.l)if self.right:self.right.v = self.vself.right.ans = self.f2(self.v, self.right.r - self.right.l)self.v = '#'# push upself.ans = self.f1(self.left.cover_seg(l, r, v), self.right.cover_seg(l, r, v))return self.ansdef inc_seg(self, l, r, v):'''将线段[left,right)增加val'''if v == 0 or l >= self.r or r <= self.l:return self.ans# self.ans = '?'if l <= self.l and r >= self.r:if self.v == '#':self.lazy_tag += velse:self.v += vself.ans += self.f2(v, self.r - self.l)return self.ansself.create_subtrees()if self.v != '#':self.left.v = self.vself.left.ans = self.f2(self.v, self.left.r - self.left.l)self.right.v = self.vself.right.ans = self.f2(self.v, self.right.r - self.right.l)self.v = '#'self.pushdown()self.ans = self.f1(self.left.inc_seg(l, r, v), self.right.inc_seg(l, r, v))return self.ansdef inc_idx(self, idx, v):'''increase idx by val'''if v == 0 or idx >= self.r or idx < self.l:return self.ansif idx == self.l == self.r - 1:self.v += vself.ans += self.f2(v, 1)return self.ansself.create_subtrees()if self.v != '#':self.left.v = self.vself.left.ans = self.f2(self.v, self.left.r - self.left.l)self.right.v = self.vself.right.ans = self.f2(self.v, self.right.r - self.right.l)self.v = '#'self.pushdown()self.ans = self.f1(self.left.inc_idx(idx, v), self.right.inc_idx(idx, v))return self.ansdef pushdown(self):if self.lazy_tag != 0:if self.left:if self.left.v != '#':self.left.v += self.lazy_tagself.left.lazy_tag = 0else:self.left.lazy_tag += self.lazy_tagself.left.ans += self.f2(self.lazy_tag, self.left.r - self.left.l)if self.right:if self.right.v != '#':self.right.v += self.lazy_tagself.right.lazy_tag = 0else:self.right.lazy_tag += self.lazy_tagself.right.ans += self.f2(self.lazy_tag, self.right.r - self.right.l)self.lazy_tag = 0def query(self, l, r):'''查询线段[right,bottom)的RMQ'''if l >= r: return 0if l <= self.l and r >= self.r:return self.ansif self.v != '#':return self.f2(self.v, min(self.r, r) - max(self.l, l))midh = self.mid_hanss = []if l < midh:anss.append(self.left.query(l, r))if r > midh:anss.append(self.right.query(l, r))return reduce(self.f1, anss)n, m = list(map(int, input().split()))
a = list(map(int, input().split()))
ls = defaultdict(list)
intervals = []
for _ in range(m):l, r = list(map(int, input().split()))intervals.append([l, r])# fix l, get many rls[l - 1].append(r - 1)if m == 0:print(max(a) - min(a))print(0)print()
else:# fix max# use segTreesf1 = lambda a, b: min(a, b)f2 = lambda a, n: asegtree = SegTree(f1, f2, 0, n, 0)# initfor i in range(n):segtree.inc_idx(i, a[i])# delete all intervals# if we later fix a maxn, we can add corresponding intervalsfor l, r in intervals:segtree.inc_seg(l - 1, r, -1)maxD, maxI = 0, 1rs = defaultdict(list)for i in range(n):# fix ifor r in ls[i]:# add intervals back(not choose)segtree.inc_seg(i, r + 1, 1)rs[r].append(i)# delete useless intervals again# means that we choose these intervalsfor l in rs[i - 1]:segtree.inc_seg(l, i, -1)# find now dd = a[i] - segtree.query(0, n)if d > maxD:maxD, maxI = d, iids = []for i, interval in enumerate(intervals):if interval[1] - 1 < maxI or interval[0] - 1 > maxI: # outsideids.append(i + 1)print(maxD)print(len(ids))print(*ids)

总结

线段树 + 全删逆向 + 剔除包含当前节点的所有区间的ls + rs算法

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codeforces：E2. Array and Segments (Hard version)【线段树 + 区间修改】

分析

暴力解法

优化

segTree solution

总结

codeforces：E2. Array and Segments (Hard version)【线段树 + 区间修改】相关推荐

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