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…

题目大意:给出一棵树,n(n<=5w)个节点,k(k<=10w)次修改,每次给定s和t,把s到t的路径上的点权+1,问k次操作后最大点权. 对于每次修改,给s和t的点权+1,给lca(s,t)和lca(s,t)的父亲的点权-1,每一个点的权就是它与它的子树权和,实际上就是树上的差分,又涨姿势了... 代码如下: uses math; type point=^rec; rec=record data:longint; next:point; end; var n,m,x,y,i,ans,fa,k…

[Luogu 3128] USACO15DEC Max Flow 最近跟 LCA 干上了- 树剖好啊,我再也不想写倍增了. 以及似乎成功转成了空格选手 qwq. 对于每两个点 S and T,求一下 LCA 顺便树上差分,最后求差分数组的前缀和并找出最大值输出就行了. (PS:最近考前训练不开 C++11,所以如果看见我写了奇怪的 define 请自动无视QAQ!) #include <algorithm> #include <cstdio> #define nullptr NUL…

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 N−1N-1N−1 pipes to transport milk between the NNN stalls in his barn (2≤N≤50,0002 \leq N \leq 50,0002≤N≤50,000), conveniently numbered 1-N1 \ldots N1-N. Each pipe connects a pair of stalls, and all stal…

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…

跟LOJ10131暗的连锁 相似,只是对于\(lca\)节点把它和父亲减一 #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a)) #define nR(a,b,c) for(regis…

题目传送门 题目描述: N个点,形成一个树状结构.有M次发放,每次选择两个点x,y对于x到y的路径上(含x,y)每个点发一袋Z类型的物品.完成所有发放后,每个点存放最多的是哪种物品. 输入格式: 第一行数字N,M接下来N-1行,每行两个数字a,b,表示a与b间有一条边再接下来M行,每行三个数字x,y,z如题 输出格式: . 样例: 样例输入: 20 508 610 618 620 107 202 1819 81 614 2016 1013 193 1417 1811 194 1115 145 1…

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…

题意:一棵树,多次给指定链上的节点加1,问最大节点权值 n个点,n-1条边很容易惯性想成一条链,幸好有样例.. 简单的树剖即可!(划去) 正常思路是树上差分,毕竟它就询问一次.. #include<iostream> #include<cstring> #include<cstdio> using namespace std; inline int rd(){ ,f=;char c; :; +c-',c=getchar(); return ret*f; } <&l…