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…

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…

题目链接:https://www.luogu.org/problemnew/show/P3128 菜 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define ll long long #define lson left, mid, rt<<1 #define rson mid+1, right, rt<<1|1 usin…

题目:https://www.luogu.org/problemnew/show/P3128 倍增求 lca 也写错了活该第一次惨WA. 代码如下: #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; ; ],dep[maxn],s[maxn],mx; bool vis[maxn]; struct N{ int to,nx…

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…

题意 题目链接 Sol 树上差分模板题 发现自己傻傻的分不清边差分和点差分 边差分就是对边进行操作,我们在\(u, v\)除加上\(val\),同时在\(lca\)处减去\(2 * val\) 点差分是对点操作,我们在\(u, v\)处加上\(val\),在\(lca\)和\(fa[lca]\)处减去\(val\) 就本题而言,属于点差分 #include<bits/stdc++.h> const int MAXN = 1e5 + 10; using namespace std; inline…