咸鱼

将树压缩成若干个有用的点for(int i=0;i<k;++i){ int f=lca(stk[r],h[i]); while(r&&id[f]<id[stk[r-1]]){ g1.add(stk[r-1],stk[r],get(stk[r],stk[r-1])); --r; } if(f^stk[r])...

#include<bits/stdc++.h>using namespace std;char buf[1<<20],*_=buf,*__=buf;#define gc() (_==__&&(__=(_=buf)+fread(buf,1,1<<20,stdin),_==__)?EOF:*_++)#define TT template&lt...

Tarjan割点low[x]:x能到达的cnt最小的节点（包括x自身节点）对于x节点，存在dfs树上x的子节点y，low[y]>=low[x]，即y能到达的最小编号节点是以x为根节点的子树，删除x之后，y与x的父节点不连通（特判根节点）void tarjan(int x){ low[x]=dfn[x]=++tarcnt; int flag=0; for(int...

好像很少有关于top数组性质的博客。for(int i=top[x];fa[i];i=top[fa[i]])可以遍历所有子树中含有x节点的重链链头节点（或叶子节点），由于重链不超过log(n)个,所以这个循环复杂度是O(logN)的。设i是重链链头节点，f是i的父节点，则i是f的轻儿子。所以，如果有操作：对于x节点，他的所有儿子均加上不同的值，且这个值只与儿子节点自身和提供的值有关，那么...

Cheap RobotYou’re given a simple, undirected, connected, weighted graph with n nodes and m edges.Nodes are numbered from 1 to n. There are exactly k centrals (recharge points), which are nodes 1,2,…...

解决图中：任意两节点（可以不连通）找到x<->y路径中边权的最小值最大，反之亦然（也可以用树剖写）给定起点，经过的路径边权有某限制下的（如小于等于某值）点权第k小（大），需要主席树。对于1：看着像二分。。对原图边权排序，生成树是直接并查集merge x，y两个节点，重构树的话会新生成一个节点，并把原来的x与y的边断开，用边权当做新节点的点权。//ff[]是并查集for...

Test for JobMr.Dog was fired by his company. In order to support his family, he must find a new job as soon as possible. Nowadays, It’s hard to have a job, since there are swelling numbers of the une...

MMSet2给定一棵n个节点的树，点编号为1…n。Q次询问，每次询问给定一个点集S，令，f(u)=max⁡v∈Sdist(u,v)f(u)=\max_{v\in S}dist(u,v)f(u)=maxv∈S​dist(u,v)你需要求出min⁡u=1…nf(u)\min_{u=1\dots n}f(u)minu=1…n​f(u)其中dist(u,v)表示树上路径(u,v)的边数。输入描述...

SuperdokuAlice and Bob are big fans of math. In particular, they are very excited about playing games that are related to numbers. Whenever they see a puzzle like Sudoku, they cannot stop themselves ...

Telephone Lines 题目描述Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his fa...