ACM-ICPC 2018 焦作赛区网络预赛 E Jiu Yuan Wants to Eat (树链剖分+线段树)
题目链接:https://nanti.jisuanke.com/t/31714
题意:给你一棵树,初始全为0,有四种操作:
1.u-v乘x 2.u-v加x 3. u-v取反 4.询问u-v的和
思路:
除去第三个操作就是很简单的树链剖分+线段树多重标记下放,所以我们只要考虑怎么维护第三个操作就好了,
由题目给的取反可知:!x = (2^64-1) - x; 但是这样维护还是很麻烦,因为这道题是对2^64取模的,我们可以
尝试把这个式子转换成只有加法和乘法的,这样就可以将其和前面两个操作一起维护降低难度。
我们可以想到: (-x)%(2^64) = (2^64-1)*x%(2^64),那么式子就可以转换成:
!x = (2^64-1)*x + (2^64-1) 这样这个式子就只有乘法和加法操作,以及一个常数,然后带入多重标记维护就好了
数组要开 unsigned long long ,题目给的模数也很特殊,2^64, 因为unsigned long long 溢出的时候相当于对2^64
取模,那么我们就不需要取模操作了,直接运算就完事了。
实现代码:
#include<bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
#define ll unsigned long long
const ll inf = ;
const int M = 2e5+; vector<int>g[M];
int n,cnt1,cnt,a[M];
int son[M],fa[M],head[M],siz[M],top[M],dep[M],tid[M];
ll sum[M<<],add[M<<],mul[M<<]; void dfs1(int u,int faz,int deep){
dep[u] = deep;
fa[u] = faz;
siz[u] = ;
for(int i = ;i < g[u].size();i++){
int v = g[u][i];
//cout<<v<<" ";
if(v != fa[u]){
dfs1(v,u,deep+);
siz[u] += siz[v];
if(son[u] == -||siz[v] > siz[son[u]])
son[u] = v;
}
}
} void dfs2(int u,int t){
top[u] = t;
tid[u] = ++cnt;
if(son[u] == -) return;
dfs2(son[u],t);
for(int i = ;i<g[u].size();i++){
int v = g[u][i];
if(v != son[u]&&v != fa[u])
dfs2(v,v);
}
} void pushup(int rt){
sum[rt] = sum[rt<<] + sum[rt<<|];
} void pushdown(int m,int rt){
add[rt<<] = add[rt<<]*mul[rt] + add[rt];
add[rt<<|] = add[rt<<|]*mul[rt]+add[rt];
mul[rt<<] = mul[rt<<]*mul[rt];
mul[rt<<|] = mul[rt<<|]*mul[rt];
sum[rt<<] = sum[rt<<]*mul[rt] + add[rt]*(m-(m>>));
sum[rt<<|] = sum[rt<<|]*mul[rt]+add[rt]*(m>>);
add[rt] = ; mul[rt] = ;
} void build(int l,int r,int rt){
add[rt] = ; mul[rt] = ;
if(l == r){
sum[rt] = ;
return;
}
mid;
build(lson); build(rson);
pushup(rt);
} void update(int L,int R,ll c,int v,int l,int r,int rt){
if(L <= l&&R >= r){
if(v == ){
sum[rt] = sum[rt]*c;
add[rt] = add[rt]*c;
mul[rt] = mul[rt]*c;
}
else if(v == ){
sum[rt] = sum[rt] + (ll)c*(r-l+);
add[rt] = add[rt]+c;
}
else if(v == ){
sum[rt] = inf*sum[rt] + (r-l+)*inf;
add[rt] = add[rt]*inf+inf;
mul[rt] *= inf;
}
return;
}
pushdown(r-l+,rt);
int m = (l + r) >> ;
if(L <= m) update(L,R,c,v,lson);
if(R > m) update(L,R,c,v,rson);
pushup(rt);
} ll query(int L,int R,int l,int r,int rt){
if(L <= l&&R >= r){
return sum[rt];
}
mid;
pushdown(r-l+,rt);
ll ret = ;
if(L <= m) ret += query(L,R,lson);
if(R > m) ret += query(L,R,rson);
return ret;
} void cover(int x,int y,ll c,int v){
int fx = top[x],fy = top[y];
while(fx!=fy){
if(dep[fx] < dep[fy]) swap(fx,fy),swap(x,y);
update(tid[fx],tid[x],c,v,,n,);
x = fa[fx];fx = top[x];
}
if(dep[x] < dep[y]) swap(x,y);
update(tid[y],tid[x],c,v,,n,);
} ll ask(int x,int y){ //求两结点路径上的权值和
int fx = top[x],fy = top[y];
ll ans = ;
while(fx != fy){
if(dep[fx] < dep[fy]) swap(fx,fy),swap(x,y);
ans += query(tid[fx],tid[x],,n,);
x = fa[fx]; fx = top[x];
}
ans += (dep[x] > dep[y])?query(tid[y],tid[x],,n,):query(tid[x],tid[y],,n,);
return ans;
} void init(){
cnt = cnt1 = ;
memset(son,-,sizeof(son));
dep[] = ; fa[] = ; siz[] = ;
} int main()
{
int x,q,op,l,r;
ll c;
while(scanf("%d",&n)!=EOF){
init();
// cout<<inf<<endl;
for(int i = ;i <= n;i ++){
scanf("%d",&x);
g[x].push_back(i);
}
dfs1(,,);
dfs2(,);
build(,n,);
scanf("%d",&q);
while(q--){
scanf("%d",&op);
if(op == ||op == ){
scanf("%d%d%llu",&l,&r,&c);
cover(l,r,c,op);
}
else if(op == ){
scanf("%d%d",&l,&r);
cover(l,r,,op);
}
else {
scanf("%d%d",&l,&r);
printf("%llu\n",ask(l,r));
}
}
for(int i = ;i <= n;i ++)
g[i].clear();
}
return ;
}
ACM-ICPC 2018 焦作赛区网络预赛 E Jiu Yuan Wants to Eat (树链剖分+线段树)的更多相关文章
- ACM-ICPC 2018 焦作赛区网络预赛 E. Jiu Yuan Wants to Eat (树链剖分-线性变换线段树)
树链剖分若不会的话可自行学习一下. 前两种操作是线性变换,模\(2^{64}\)可将线段树全部用unsigned long long 保存,另其自然溢出. 而取反操作比较不能直接处理,因为其模\(2^ ...
- ACM-ICPC 2018 焦作赛区网络预赛- G:Give Candies(费马小定理,快速幂)
There are N children in kindergarten. Miss Li bought them NNN candies. To make the process more inte ...
- ACM-ICPC 2018 焦作赛区网络预赛- L:Poor God Water(BM模板/矩阵快速幂)
God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him t ...
- ACM-ICPC 2018 焦作赛区网络预赛
这场打得还是比较爽的,但是队友差一点就再过一题,还是难受啊. 每天都有新的难过 A. Magic Mirror Jessie has a magic mirror. Every morning she ...
- ACM-ICPC 2018 焦作赛区网络预赛J题 Participate in E-sports
Jessie and Justin want to participate in e-sports. E-sports contain many games, but they don't know ...
- ACM-ICPC 2018 焦作赛区网络预赛 K题 Transport Ship
There are NN different kinds of transport ships on the port. The i^{th}ith kind of ship can carry th ...
- ACM-ICPC 2018 焦作赛区网络预赛 L 题 Poor God Water
God Water likes to eat meat, fish and chocolate very much, but unfortunately, the doctor tells him t ...
- ACM-ICPC 2018 焦作赛区网络预赛 I题 Save the Room
Bob is a sorcerer. He lives in a cuboid room which has a length of AA, a width of BB and a height of ...
- ACM-ICPC 2018 焦作赛区网络预赛 H题 String and Times(SAM)
Now you have a string consists of uppercase letters, two integers AA and BB. We call a substring won ...
随机推荐
- maven 插
一.maven插件元素 <?xml version="1.0" encoding="utf-8"?> <plugin> <!--插 ...
- JS-隐士类型转换‘1’+1、‘1’-1、++‘1’为什么不一样?
当 x=’1’时,x+1x-1+x-x++xtypeof(x+1)typeof(x-1)typeof(+x)typeof(-x)typeof(++x) 的结果分别是多少? 答案: x+1 //’11’ ...
- 扩展 WPF 动画类
原文:扩展 WPF 动画类 扩展 WPF 动画类 Charles ...
- asp.net mvc 实现上传文件带进度条
本文乃是博主早期写的,此种思路虽然实现了,但固然不是最好的,仅做参考学习. 可以用js onprogress .fileinput .webuploader.jq ajaxsubmit等实现 思路:a ...
- Jmeter(非GUI模式)教程
前言 使用非 GUI 模式,即命令行模式运行 JMeter 测试脚本能够大大缩减所需要的系统资源.优点如下:1.节约系统资源:无需启动界面,节约系统资源 2.便捷快速:仅需启动命令行,输入命令便可执行 ...
- Codeforces Round #504 (rated, Div. 1 + Div. 2, based on VK Cup 2018 Final)-C-Bracket Subsequence
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> ...
- 【2016.3.18】作业 VS2015安装&单元测试(2)
- Linux内核第七节 20135332武西垚
预处理.编译.链接和目标文件的格式 可执行程序是怎么得来的 以C语言为例,c代码经过编译器的预处理,编译成汇编代码,由汇编器编译成目标代码,再链接成可执行文件,由操作系统加载到cpu里来执行. (截图 ...
- PolarCode
什么是polar code极化码 为了实现可靠的信号传输,编码学家在过去的半个多世纪提出多种纠错码技术如里所码(RS码).卷积码,Turbo码等,并在各种通信系统中取得了广泛的应用.但是以往所有实用的 ...
- 小学四则运算APP 第二阶段冲刺-第五天
团队成员:陈淑筠.杨家安.陈曦 团队选题:小学四则运算APP 第二次冲刺阶段时间:11.29~12.09 本次发布的是判断题代码,已经实现部分功能,,但是美中不足的是判断错误 panduanset.j ...