传送门

因为某些原因,所以我就去学了 $LCT$ 维护直径, $LCT$ 维护直径我上一个博客讲得很详细了:传送门

这里维护虚儿子用的是 $multiset$ ,没写可删堆

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;
inline int read()
{
int x=,f=; char ch=getchar();
while(ch<''||ch>'') { if(ch=='-') f=-; ch=getchar(); }
while(ch>=''&&ch<='') { x=(x<<)+(x<<)+(ch^); ch=getchar(); }
return x*f;
}
const int N=8e5+;
const ll INF=1e18;
int n,m;
struct edge {
int x,y;
}e[N];
namespace LCT {
int c[N][],fa[N],sz[N],val[N];
ll sum[N],lmx[N],rmx[N],mxs[N];
bool rev[N];
multiset <ll> H[N],P[N];
inline void ins(int u,int v) { H[u].insert(lmx[v]); P[u].insert(mxs[v]); }
inline void del(int u,int v) { H[u].erase(H[u].find(lmx[v])); P[u].erase(P[u].find(mxs[v])); }
inline ll fir(multiset <ll> &S) { return S.size() ? *S.rbegin() : -INF; }
inline ll sec(multiset <ll> &S) { return S.size()> ? *(++S.rbegin()) : -INF; }
inline void pushup(int x)
{
int &lc=c[x][],&rc=c[x][];
sum[x]=sum[lc]+sum[rc]+val[x];
ll t=max(0ll,fir(H[x])), L=max(t,rmx[lc])+val[x], R=max(t,lmx[rc])+val[x];
lmx[x]=max(lmx[lc], sum[lc]+R ); rmx[x]=max(rmx[rc], sum[rc]+L );
mxs[x]=max( max( rmx[lc]+R , lmx[rc]+L ) , max(mxs[lc],mxs[rc]) );
mxs[x]=max(mxs[x],fir(P[x])); mxs[x]=max(mxs[x], t+val[x] );
mxs[x]=max(mxs[x], t+val[x]+sec(H[x]) );
}
inline void pushdown(int x)
{
if(!x||!rev[x]) return;
int &lc=c[x][],&rc=c[x][]; rev[x]=;
swap(lc,rc); swap(lmx[x],rmx[x]);
if(lc) rev[lc]^=;
if(rc) rev[rc]^=;
}
inline void rever(int x) { rev[x]=; pushdown(x); }
inline bool noroot(int x) { return (c[fa[x]][]==x)|(c[fa[x]][]==x); }
inline void rotate(int x)
{
int y=fa[x],z=fa[y],d=(c[y][]==x);
if(noroot(y)) c[z][c[z][]==y]=x;
fa[x]=z; fa[y]=x; fa[c[x][d^]]=y;
c[y][d]=c[x][d^]; c[x][d^]=y;
pushup(y);
}
void push_rev(int x)
{
if(noroot(x)) push_rev(fa[x]);
else pushdown(x);
pushdown(c[x][]); pushdown(c[x][]);
}
inline void splay(int x)
{
push_rev(x);
while(noroot(x))
{
int y=fa[x],z=fa[y];
if(noroot(y))
(c[y][]==x ^ c[z][]==y) ? rotate(x) : rotate(y);
rotate(x);
} pushup(x);
}
inline void access(int x)
{
for(int y=;x;y=x,x=fa[x])
{
splay(x); if(y) del(x,y);
if(c[x][]) ins(x,c[x][]);
c[x][]=y; pushup(x);
}
}
inline void makeroot(int x) { access(x); splay(x); rever(x); }
inline int findroot(int x)
{
access(x); splay(x); pushdown(x);
while(c[x][]) x=c[x][],pushdown(x);
splay(x); return x;
}
inline void split(int x,int y) { makeroot(x); access(y); splay(y); }
inline void link(int x,int y)
{
makeroot(x); if(findroot(y)==x) return;
makeroot(y); fa[x]=y; ins(y,x); pushup(y);
}
inline void cut(int x,int y)
{
makeroot(x);
if(findroot(y)!=x||fa[y]!=x||c[y][]) return;
fa[y]=c[x][]=; pushup(x);
}
inline void Link(int k) { link(e[k].x,n+k); link(e[k].y,n+k); }
inline void Cut(int k) { cut(e[k].x,n+k); cut(e[k].y,n+k); }
inline ll query(int x) { access(x); splay(x); return rmx[x]; }
}
int main()
{
n=read();
for(int i=;i<n;i++)
e[i].x=read(),e[i].y=read(),LCT::val[n+i]=read();
for(int i=;i<n;i++) LCT::Link(i);
m=read(); char s[]; int a;
for(int i=;i<=m;i++)
{
scanf("%s",s); a=read();
if(s[]=='Q') printf("%lld\n",LCT::query(a));
else LCT::Cut(a),LCT::val[n+a]=read(),LCT::Link(a);
}
return ;
}

The Preliminary Contest for ICPC Asia Shanghai 2019 A. Lightning Routing I的更多相关文章

  1. The Preliminary Contest for ICPC Asia Shanghai 2019 C Triple(FFT+暴力)

    The Preliminary Contest for ICPC Asia Shanghai 2019 C Triple(FFT+暴力) 传送门:https://nanti.jisuanke.com/ ...

  2. The Preliminary Contest for ICPC Asia Shanghai 2019

    传送门 B. Light bulbs 题意: 起初\(n\)个位置状态为\(0\),\(m\)次操作,每次操作更换区间状态:\(0\)到\(1\),\(1\)到\(0\). 共有\(T,T\leq 1 ...

  3. The Preliminary Contest for ICPC Asia Shanghai 2019 C. Triple

    [传送门] FFT第三题! 其实就是要求有多少三元组满足两短边之和大于等于第三边. 考虑容斥,就是枚举最长边,另外两个数组里有多少对边之和比它小,然后就是 $n^3$ 减去这个答案. 当 $n \le ...

  4. 01背包方案数(变种题)Stone game--The Preliminary Contest for ICPC Asia Shanghai 2019

    题意:https://nanti.jisuanke.com/t/41420 给你n个石子的重量,要求满足(Sum<=2*sum<=Sum+min)的方案数,min是你手里的最小值. 思路: ...

  5. 给定进制下1-n每一位数的共享(Digit sum)The Preliminary Contest for ICPC Asia Shanghai 2019

    题意:https://nanti.jisuanke.com/t/41422 对每一位进行找循环节规律就行了. #define IOS ios_base::sync_with_stdio(0); cin ...

  6. The Preliminary Contest for ICPC Asia Shanghai 2019 L. Digit sum

    题目:https://nanti.jisuanke.com/t/41422 思路:预处理 #include<bits/stdc++.h> using namespace std; ][]= ...

  7. The Preliminary Contest for ICPC Asia Shanghai 2019 J. Stone game

    题目:https://nanti.jisuanke.com/t/41420 思路:当a(a∈S′)为最小值 如果Sum(S′)−a≤Sum(S−S′)成立 那么(∀t∈S′,Sum(S′)−t≤Sum ...

  8. The Preliminary Contest for ICPC Asia Shanghai 2019 D. Counting Sequences I

    题目:https://nanti.jisuanke.com/t/41412思路:dfs           先取ai>2  2^12>3000 因此至多取11个 其余用1补        ...

  9. The Preliminary Contest for ICPC Asia Shanghai 2019 B. Light bulbs

    题目:https://nanti.jisuanke.com/t/41399 思路:差分数组 区间内操作次数为奇数次则灯为打开状态 #include<bits/stdc++.h> using ...

随机推荐

  1. Luogu P4707 重返现世 (拓展Min-Max容斥、DP)

    题目链接 https://www.luogu.org/problem/P4707 题解 最近被神仙题八连爆了-- 首先Min-Max容斥肯定都能想到,问题是这题要用一个扩展版的--Kth Min-Ma ...

  2. 利用H5缓存机制实现点击按钮第一次与之后再点击分别跳转不同页面

    昨天碰到这样一个需求,要求点击按钮第一次跳转到a页面,之后再点击它就跳转到b页面.这个问题我首先就想到了利用H5的缓存sessionstorage来实现,SessionStorage用于本地存储一个会 ...

  3. SELECT list is not in GROUP BY clause and contains nonaggregated

    安装了mysql5.7,用group by 查询时抛出如下异常 SQLSTATE[42000]: Syntax error or access violation: 1055 Expression # ...

  4. python模块------pyautogui

    安装 pip install pyautogui 基本使用 查询 screenWidth, screenHeight = pyautogui.size() # 屏幕尺寸 mouseX, mouseY ...

  5. windows+VS+Openjdk12 搭建debug环境

    openjdk12 下载地址: http://jdk.java.net/archive/ https://download.java.net/java/GA/jdk12/33/GPL/openjdk- ...

  6. leetcode 116填充每个节点的下一个右侧节点指针

    time O(n) ,sapce O(n) /* // Definition for a Node. class Node { public: int val; Node* left; Node* r ...

  7. 浏览器端-W3School-HTML:HTML DOM Select 对象

    ylbtech-浏览器端-W3School-HTML:HTML DOM Select 对象 1.返回顶部 1. HTML DOM Select 对象 Select 对象 Select 对象代表 HTM ...

  8. 用xmmp+openfire+smack搭建简易IM实现

    功能实现:注册,登录,单聊表情,文本,图片,语音的发送接收,添加好友,删除好友,查找好友,修改密码,消息提醒设置,获取离线消息等功能 1.前期准备 1.下载opnefire软件:https://www ...

  9. 解决ssh链接慢问题

    1,ssh -v root@ip 2,查看哪里卡住了 一般情况就是卡在debug1: SSH2_MSG_SERVICE_ACCEPT received 3,如果是上面卡住了 修改/etc/ssh/ss ...

  10. Ubuntu系统下各种报错杂烩(持续更新)

    在Github时报Permanently added the RSA host key for IP address '192.30.252.128' to the list of known hos ...