传送门 这是个假的最大流,其实是一个用树剖+线段树就能解决的事情 题目中的道路会对路径上的造成压力,最后询问最大的压力 其实就等价于对每条路径上的点加上 1 的权值,并且最后询问整个树中的最大值 然后树剖+最大值线段树裸题,完事,莫得别的问题了. Code: #include <iostream> #include <cstdlib> #include <cstdio> #include <vector> #define ls ( rt << 1…

因为徐州现场赛的G是树上差分+组合数学,但是比赛的时候没有写出来(自闭),背锅. 会差分数组但是不会树上差分,然后就学了一下. 看了一些东西之后,对树上差分写一点个人的理解: 首先要知道在树上,两点之间只有一条路径.树上差分就是在树上用差分数组,因为是在树上的操作,所以要用到lca,因为对于两点a,b,从a到b这一条链就是a-->lca(a,b)-->b,这是一条链. 其次,树上差分的两种操作:一种是对点权的,另一种是对边权的. 对于点权: 在树上将路径的起点a+1和终点b+1,lca(a,b…

P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of N-1N−1 pipes to transport milk between the NN stalls in his barn (2 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1 \ldots N1…N. Each pipe connects a pair of stalls,…

P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via path…

链接一下题目:luoguP3128 [USACO15DEC]最大流Max Flow(树上差分板子题) 如果没有学过树上差分,抠这里(其实很简单的,真的):树上差分总结 学了树上差分,这道题就极其显然了,不就是把每一条运输路线差分进去,那就是板子了啊. 树上差分还是很有用的,比较容易写,这种询问很少的题目去敲那么长(还容易出玄学错误)的树剖很浪费,用树上差分就很快了!(//...微笑...\\) 上一波代码: #include<iostream> #include<cstdlib>…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

题目描述 Farmer John has installed a new system of N-1N−1 pipes to transport milk between the NN stalls in his barn (2 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1 \ldots N1…N. Each pipe connects a pair of stalls, and all stalls are connected t…

讲解: https://rpdreamer.blog.luogu.org/ci-fen-and-shu-shang-ci-fen #include <bits/stdc++.h> #define read read() #define up(i,l,r) for(register int i = (l);i <= (r);i++) #define down(i,l,r) for(register int i = (l);i >= (r);i--) #define traversal…

思路 这个题哪里有那么费脑筋 我们可以树链剖分嘛LCT昨天学的时候睡着了,不是太会 两遍dfs+一个5行的BIT 其实树链剖分学好了对倍增和LCT理解上都有好处 一条路径上的修改 由于一条剖出来的链是连续的,我们要选择数据结构维护 不过这里不用维护太多东西,只是区间+1 我们可以选择常数小,好写的树状数组(从50行的线段树变成5行的 BIT) 而且使得\(O(nlog_{2})\)的算法跑的并不慢 具体就是用差分思想,修改区间[L,R]时 $[1,R] +1 $ \([1,L-1] -1\)达到…