原文地址:http://www.cnblogs.com/GXZlegend/p/6832263.html 题目描述 The cows have once again tried to form a startup company, failing to remember from past experience that cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organi…
题意 题目链接 Sol 线段树合并板子题 #include<bits/stdc++.h> using namespace std; const int MAXN = 400000, SS = MAXN * 21; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >=…
4756: [Usaco2017 Jan]Promotion Counting Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 305  Solved: 217[Submit][Status][Discuss] Description The cows have once again tried to form a startup company, failing to remember from past experience t hat cow…
调半天原来是dsu写不熟 Description The cows have once again tried to form a startup company, failing to remember from past experience t hat cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organize t he company as a tree, with…
传送门 此题很有意思,有多种解法 1.用天天爱跑步的方法,进入子树的时候ans-query,出去子树的时候ans+query,query可以用树状数组或线段树来搞 2.按dfs序建立主席树 3.线段树的合并 前两个都会,于是学习一下线段树的合并.. 道理用文字解释不清...直接看代码就能看懂.. 可以脑补出,合并的操作复杂度是logn的,总时间复杂度nlogn #include <cstdio> #include <cstring> #include <iostream>…
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=4756 [题解] dsu on tree,树状数组直接上 O(nlog^2n) # include <vector> # include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.…
题目描述 The cows have once again tried to form a startup company, failing to remember from past experience t hat cows make terrible managers!The cows, conveniently numbered 1…N1…N (1≤N≤100,000), organize t he company as a tree, with cow 1 as the preside…
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=4756 线段树合并裸题.那种返回 int 的与传引用的 merge 都能过.不知别的题是不是这样. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ,M=N**; int n,m,a[N],tp[N],rt[N],…
HDU 5862 Counting Intersections(离散化+树状数组) 题目链接http://acm.split.hdu.edu.cn/showproblem.php?pid=5862 Description Given some segments which are paralleled to the coordinate axis. You need to count the number of their intersection. The input data guarant…
花了近5个小时,改的乱七八糟,终于A了. 一个无限数列,1,2,3,4,...,n....,给n个数对<i,j>把数列的i,j两个元素做交换.求交换后数列的逆序对数. 很容易想到离散化+树状数组,但是发现那些没有交换的数也会产生逆序对数,但我没有算. 经明神提示, 把没有用到的数字段化成点.然后用树状数组算一下就好了. 然后我用一个数组记录每个点的长度.比如 <1,2><5,6>,1,2,3,4,5,6只有1,2,5,6用到了,那么离散化为1,2,3,4,5,f[1]=…