HDU 4010.Query on The Trees 解题报告
题意:
给出一颗树,有4种操作:
1、如果x和y不在同一棵树上则在xy连边
2、如果x和y在同一棵树上并且x!=y则把x换为树根并把y和y的父亲分离
3、如果x和y在同一棵树上则x到y的路径上所有的点权值+w
4、如果x和y在同一棵树上则输出x到y路径上的最大值
动态树入门题:
#include <iostream>
#include <cstdio>
using namespace std; const int MAXN = 333333; struct node {
int val, max, inc;
bool rev;
node *par, *Ch[2];
} dt[MAXN], *NIL = dt; struct LinkcutTree {
inline void _inc (node * x, int inc) {
if (x == NIL) return;
x->inc += inc, x->val += inc, x->max += inc;
}
inline void clear (node *const x) {
if (x == NIL) return ;
if (x->inc) {
_inc (x->Ch[0], x->inc);
_inc (x->Ch[1], x->inc);
x->inc = 0;
}
if (x->rev) {
swap (x->Ch[0], x->Ch[1]);
x->Ch[0]->rev ^= 1;
x->Ch[1]->rev ^= 1;
x->rev = 0;
}
}
inline void update (node * x) {
clear (x);
clear (x->Ch[0]), clear (x->Ch[1]);
x->max = max (x->val, max (x->Ch[0]->max, x->Ch[1]->max) );
}
void Rotate (node *x) {
node *p = x->par, *g = p->par;
clear (p),clear (x);
int c = p->Ch[0] == x; //0左旋,1右旋
p->Ch[c ^ 1] = x->Ch[c];
if (x->Ch[c] != NIL) x->Ch[c]->par = p;
x->par = g;
if (g->Ch[0] == p) g->Ch[0] = x; else
if (g->Ch[1] == p) g->Ch[1] = x;
x->Ch[c] = p;
p->par = x;
update (p);
}
void Splay (node *x) {
if(x==NIL) return ;
while (x->par != NIL && (x->par->Ch[0] == x || x->par->Ch[1] == x) ) {
if (x->par != NIL)
Rotate (x);
else {
node *p = x->par, *g = p->par;
if ( (g->Ch[1] == p) == (p->Ch[1] == x) )
Rotate (p), Rotate (x);
else
Rotate (x), Rotate (x);
}
}
update (x);
}
node *Access (node *u) {
node *v = NIL;
for (; u != NIL; u = u->par) {
Splay (u);
u->Ch[1] = v;
update (v = u);
}
return v;
}
node *getroot (node *x) {
for (x = Access (x); clear (x), x->Ch[0] != NIL; x = x->Ch[0]);
return x;
}
inline void evert (node *x) {
Access (x)->rev ^= 1;
Splay (x);
}
inline void link (node *x, node *y) {
evert (x);
x->par = y;
Access (x);
}
inline void cut (node *x, node *y) {
evert (x);
Access (y);
Splay (y);
y->Ch[0]->par = NIL;
y->Ch[0] = NIL;
update (y);
}
inline int query (node *x, node *y) {
evert (x);
Access (y), Splay (y);
return y->max;
}
inline void modify (node *x, node *y, int w) {
evert (x);
Access (y), Splay (y);
_inc (y, w);
}
} LCT;
int n, m;
int main() {
while (scanf ("%d", &n) != EOF) {
for (int i = 0; i <= n; i++) {
dt[i].val = dt[i].max = dt[i].inc = 0;
dt[i].par = dt[i].Ch[0] = dt[i].Ch[1] = NIL;
dt[i].rev = 0;
}
for (int i = 1, x, y; i < n; ++i) {
scanf ("%d %d", &x, &y);
LCT.link (dt + x, dt + y);
}
for (int i = 1, x; i <= n; ++i) {
scanf ("%d", &x);
node *tem = dt + i;
LCT.Splay (tem);
tem->val = tem->max = x;
LCT.update (tem);
}
scanf ("%d", &m);
for (int i = 0, op, x, y, z; i < m; ++i) {
scanf ("%d%d%d", &op, &x, &y);
switch (op) {
case 1:
if (LCT.getroot (dt + x) == LCT.getroot (dt + y) ) printf ("-1\n");
else LCT.link (dt + x, dt + y); break;
case 2:
if (x == y || LCT.getroot (dt + x) != LCT.getroot (dt + y) ) printf ("-1\n");
else LCT.cut (dt + x, dt + y); break;
case 3:
scanf ("%d", &z);
if (LCT.getroot (dt + y) != LCT.getroot (dt + z) ) printf ("-1\n");
else LCT.modify (dt + y, dt + z, x); break;
case 4:
if (LCT.getroot (dt + x) != LCT.getroot (dt + y) ) printf ("-1\n");
else printf ("%d\n", LCT.query (dt + x, dt + y) ); break;
}
}
putchar (10);
}
}
HDU 4010.Query on The Trees 解题报告的更多相关文章
- 动态树(LCT):HDU 4010 Query on The Trees
Query on The Trees Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Othe ...
- HDU 4010 Query on The Trees (动态树)(Link-Cut-Tree)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4010 题意; 先给你一棵树,有 \(4\) 种操作: 1.如果 \(x\) 和 \(y\) 不在同一 ...
- HDU 4010 Query on The Trees(动态树)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4010 题意:一棵树,四种操作: (1)若x和y不在一棵树上,将x和y连边: (2)若x和y在一棵树上, ...
- HDU 4010 Query on The Trees
Problem Description We have met so many problems on the tree, so today we will have a query problem ...
- HDU 4010 Query on The Trees(动态树LCT)
Problem Description We have met so many problems on the tree, so today we will have a query problem ...
- HDU 4010 Query on The Trees(动态树)
题意 给定一棵 \(n\) 个节点的树,每个点有点权.完成 \(m\) 个操作,操作四两种,连接 \((x,y)\) :提 \(x\) 为根,并断 \(y\) 与它的父节点:增加路径 \((x,y)\ ...
- hdu 4010 Query on The Trees LCT
支持:1.添加边 x,y2.删边 x,y3.对于路径x,y上的所有节点的值加上w4.询问路径x,y上的所有节点的最大权值 分析:裸的lct...rev忘了清零死循环了两小时... 1:就是link操作 ...
- HDOJ 4010 Query on The Trees LCT
LCT: 分割.合并子树,路径上全部点的点权添加一个值,查询路径上点权的最大值 Query on The Trees Time Limit: 10000/5000 MS (Java/Others) ...
- 【LeetCode】951. Flip Equivalent Binary Trees 解题报告(Python)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcod ...
随机推荐
- Distinct Subsequences——Leetcode
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- 定时器NSTimer的用法
//时间间隔 NSTimeInterval activeTimeInterval = NETWORK_SEND_ACTIVE_TIME; NSTimeInterval othe ...
- ACM1228_STL的应用
#include<iostream> #include<string> #include<map> using namespace std; map<stri ...
- 编译android的linux kernel goldfish
https://source.android.com/source/building-kernels.html $ export PATH=/home/hzh/oldhome/learn/androi ...
- 第十七章、程序管理与 SELinux 初探
---恢复内容开始--- 什么是程序 (process) 在 Linux 底下所有的命令与你能够进行的动作都与权限有关, 而系统依据UID/GID以及文件的属性相关性判定你的权限!在 Linux 系统 ...
- 你能在windows上创建一个叫做AUX的文件夹吗?
Windows的文件名不能有如下这些特殊符号,这个大家都比较熟悉了. < (less than) > (greater than) : (colon) " (double quo ...
- javascript从入门到精通(二)
第二章.数据结构 JavaScript脚本语言的数据结构包括:标识符.关键字.常量.变量等. 标识符:就是一个名称.在JavaScript用来命名变量和函数或者用作JavaScript代码中某些循环的 ...
- windows 下删除.svn文件
Windows Registry Editor Version 5.00 [HKEY_LOCAL_MACHINE\SOFTWARE\Classes\Folder\shell\DeleteSVN] @= ...
- [RxJS] Combination operator: zip
CombineLatest and withLatestFrom are both AND-style combination operators. In this lesson, we will l ...
- IOS-tableViewCell重用机制
IOS-tableViewCell重用机制 首先介绍tableViewCell原生cell的使用和重用机制 使用dequeueReusableCellWithIdentifier方法,这个方法会从缓存 ...