Heavy Light Decomposition
Note
1.DFS1
mark all the depth
mark fathers
mark the heavy/light children
mark the size of each subtree
void dfs1(long long pos, long long f, long long depth){
dep[pos] = depth;
fat[pos] = f;
sz[pos] = 1;
long long maxi = -1;
for (long long v : adj[pos]){
if (v == f) continue;
dfs1(v,pos,depth+1);
sz[pos]+=sz[v];
if (maxi<sz[v]) {maxi = sz[v];son[pos] = v;}
}
}
2.DFS2
mark the of the members in the base array
record the top of each node
traverse heavy son first then light son
mark the heavy son along the path
give the base array a value
record the id of each node
void dfs2(long long pos, long long top_pos){
id[pos] = ++cnt;
wt[cnt] = num[pos];
top[pos] = top_pos;
if (!son[pos]) return;
dfs2(son[pos],top_pos);
for (long long v : adj[pos]){
if (v==fat[pos] || v==son[pos]) continue;
dfs2(v,v);
}
}
3.make_tree
make segment tree base on the base array
void make_tree(int way, int l, int r){
if (l==r) {seg[way] = wt[l]%p;return;}
int mid = (l+r)/2;
make_tree(way*2,l,mid);
make_tree(way*2+1,mid+1,r);
seg[way] = (seg[way*2]+seg[way*2+1])%p;
}
//push function
void push(int way,int lenn){
lazy[way*2]+=lazy[way];
lazy[way*2+1]+=lazy[way];
seg[way*2]+=lazy[way]*(lenn-(lenn>>1));
seg[way*2+1]+=lazy[way]*(lenn>>1);
seg[way*2]%=p;
seg[way*2+1]%=p;
lazy[way]=0;
}
4.query_up
query base on the segment tree
int query_up(int way, int l, int r, int qlow, int qhigh){
push(way,l,r);
if (qlow<= l && r<=qhigh) return seg[way]%p;
if (l>qhigh || r<qlow) return 0;
int mid = (l+r)/2;
return (query_up(way*2,l,mid,qlow,qhigh) + query_up(way*2+1,mid+1,r,qlow,qhigh))%p;
}
5.query
1) check if they are on the same chain. if Not, add the distance from a node to the top of the chain, and move up
2) repeat step 1 until they are on the same chain
3) query the distance between the nodes based on the base array
int query(int x, int y){
int ans = 0;
while(top[x]!=top[y]){
if (dep[top[x]]<dep[top[y]]) swap(x,y);
ans = (ans+query_up(1,1,n,id[top[x]],id[x]))%p;
x = fat[top[x]];
}
if (dep[x]>dep[y]) swap(x,y);
ans = (ans + query_up(1,1,n,id[x],id[y]))%p;
return ans;
}
6.query_son
query the id[x] plus its size -1 the subtree of a node is consecutive in the base array
int qSon(int x){
return query_up(1,1,n,id[x],id[x]+sz[x]-1);
}
7.update
update normal segment tree (use id as substitution)
void update(int way, int l, int r, int qlow, int qhigh, int val){
push(way,l,r);
if (qlow<=l && r<=qhigh) {
lazy[way] += val;
push(way,l,r);
return;
}
if (l>qhigh || r<qlow) return;
int mid = (l+r)/2;
update(way*2,l,mid,qlow,qhigh,val);
update(way*2+1,mid+1,r,qlow,qhigh,val);
seg[way] = (seg[way*2]+seg[way*2+1])%p;
}
8.update chain
1) check if they are on the same chain. If not, update the chain, and go to the father of the chain head
2) repeat until they are on the same chain
3) update the id of the nodes on the segment tree
void upPath(int x, int y, int val){
while(top[x]!=top[y]){
if (dep[top[x]]<dep[top[y]]) swap(x,y);
update(1,1,n,id[top[x]],id[x],val);
x = fat[top[x]];
}
if (dep[x]>dep[y]) swap(x,y);
update(1,1,n,id[x],id[y],val);
}
9.update_son
update id[x] plus its size -1
void upSon(int x, int val){
update(1,1,n,id[x],id[x]+sz[x]-1,val);
}
optional: LCA
1) check if the two nodes are on the same chain
2) if not, find the one with a deeper head, move it up
3) repeat until they are in the same chain
3) return the one with higher depth
int LCA (int x , int y ) {
int fx = top[x] , fy = top[y] ;
while(fx!=fy) {
if(level[fx]< level[fy]) swap(x,y) , swap(fx ,fy) ;
x = fat[fx] ; fx = top[x];
}
if( level[x] > level[y] ) swap(x,y) ;
return x ;
}
Note to myself
Segment tree is able to support different data structure
Heavy Light Decomposition的更多相关文章
- 树链剖分I 原理
树链剖分(Heavy Light Decomposition, HLD)是一种将对[树上两点间的路径]上[边或点]的[修改与查询]转化到[序列]上来处理的方法. 目的:将树的边或点转化到一个线性结构( ...
- 神奇的树上启发式合并 (dsu on tree)
参考资料 https://www.cnblogs.com/zhoushuyu/p/9069164.html https://www.cnblogs.com/candy99/p/dsuontree.ht ...
- HDU 5111 Alexandra and Two Trees 树链剖分 + 主席树
题意: 给出两棵树,每棵树的节点都有一个权值. 同一棵树上的节点的权值互不相同,不同树上节点的权值可以相同. 要求回答如下询问: \(u_1 \, v_1 \, u_2 \, v_2\):询问第一棵树 ...
- [HNOI2018]毒瘤
Description 从前有一名毒瘤. 毒瘤最近发现了量产毒瘤题的奥秘.考虑如下类型的数据结构题:给出一个数组,要求支持若干种奇奇怪怪的修改操作(比如区间加一个数,或者区间开平方),并支持询问区间和 ...
- Wolfycz的娱乐赛题解
现在不会放题解的!比赛完了我会把题解放上来的 祝大家玩的愉快~ 等会,cnblogs不会显示更新时间?我禁赛我自己 UPD:2018.12.15 欢迎大家爆踩标程- painting 我们考虑转化题意 ...
- [ZJOI2011]道馆之战
Description 口袋妖怪(又名神奇宝贝或宠物小精灵)红/蓝/绿宝石中的水系道馆需要经过三个冰地才能到达馆主的面前,冰地中的每一个冰块都只能经过一次.当一个冰地上的所有冰块都被经过之后,到下一个 ...
- codechef Heavy-light Decompositions
Heavy-light Decompositions Problem Code: HLDOTSSubmit All submissions for this problem are available ...
- ACM/ICPC 之 拓扑排序-反向(POJ3687)
难点依旧是题意....需要反向构图+去重+看题 POJ3687-Labeling Balls 题意:1-N编号的球,输出满足给定约束的按原编号排列的重量序列,如果有多组答案,则输出编号最小的Ball重 ...
- poj1013
题目大意:假造的银币 Sally Jones有一些游客给的银币,但是只有11枚是真正的银币(有一枚是假的),从颜色和大小是无法区分真比还是假币的,但是它的重量和真币是不同的,Sally Jones它是 ...
随机推荐
- eclipse中svn重新设置账户
查看svn版本:windows > preference > Team > SVN 1.如果svn插件是svnkit版 只需找到.keyring文件,一般目录是:eclipse安装目 ...
- 实验吧-隐写术-男神一般都很低调很低调的!!(stegsolve->Image Combiner + DES加密)
先介绍一下DES加密:(也可参考https://blog.csdn.net/zz_Caleb/article/details/87016017,第14个) 1)对称加密,参考:对称加密和非对称加密 2 ...
- TD信息通(无课表)使用体验
首先,在注册账户的时候,TD信息通还是比较严谨的.用户名字符数.密码字符数.邮箱格式等都有要求,我认为,这对App的长远发展来说,是很重要的一个细节.而且,在登陆之前,会有一项关于是否自动登陆的选择, ...
- python语法技巧
目录: 7.python break continue用法 1.三元表达式 2.用列表推导式来取代map和filter 3.函数式编程 4.fluent python 阅读笔记 5.enum库用法 6 ...
- select、poll和epoll机制
一.参考网址 1.select函数及fd_set介绍 2.linux select 函数和 fd_set 用法 2.select.poll和epoll的区别 3.利用select实现IO多路复用TCP ...
- docker - how do you disable auto-restart on a container?
https://stackoverflow.com/questions/37599128/docker-how-do-you-disable-auto-restart-on-a-container 9 ...
- day23(023-递归练习)
23.01_File类递归练习(统计该文件夹大小) * 需求:1,从键盘接收一个文件夹路径,统计该文件夹大小(字节?) * * 从键盘接收一个文件夹路径 * 1,创建键盘录入对象 * 2,定义一个无限 ...
- Linux学习打卡20200214
- CF940F Machine Learning(带修莫队)
首先显然应该把数组离散化,然后发现是个带修莫队裸题,但是求mex比较讨厌,怎么办?其实可以这样求:记录每个数出现的次数,以及出现次数的出现次数.至于求mex,直接暴力扫最小的出现次数的出现次数为0的正 ...
- 抓DHCP客户端ip脚本
cat testnew.sh #!/bin/bash catch_ip (){Ip=`sudo nmap -sP 192.168.1.0/24 |grep -i -B2 $mac|grep Nmap ...