hdu 3966 Aragorn's Story : 树链剖分 O(nlogn)建树 O((logn)²)修改与查询
/**
problem: http://acm.hdu.edu.cn/showproblem.php?pid=3966
裸板
**/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<vector>
using namespace std; const int MAXN = ; template <typename T>
class SegmentTree {
private:
struct Node {
int left, right;
T sum, lazy;
} node[MAXN << ];
T data[MAXN];
void pushUp(int root) {
node[root].sum = node[root << ].sum + node[root << | ].sum;
}
void pushDown(int root) {
if(node[root].left == node[root].right) return;
int lson = root << ;
int rson = root << | ;
node[lson].sum += node[root].lazy * (node[lson].right - node[lson].left + );
node[rson].sum += node[root].lazy * (node[rson].right - node[rson].left + );
node[lson].lazy += node[root].lazy;
node[rson].lazy += node[root].lazy;
node[root].lazy = ;
}
public:
void build(int left, int right, int root = ) {
node[root].left = left;
node[root].right = right;
node[root].lazy = ;
if(left == right) {
node[root].sum = data[left];
} else {
int mid = (left + right) >> ;
build(left, mid, root << );
build(mid + , right, root << | );
pushUp(root);
}
}
void update(int left, int right, T value, int root = ) {
int lson = root << ;
int rson = root << | ;
node[root].sum += value * (right - left + );
if(node[root].left == left && node[root].right == right) {
node[root].lazy += value;
return ;
}
if(left >= node[rson].left) {
update(left, right, value, rson);
} else if(right <= node[lson].right) {
update(left, right, value, lson);
} else {
update(left, node[lson].right, value, lson);
update(node[rson].left, right, value, rson);
}
}
T query(int left, int right, int root = ) {
int lson = root << ;
int rson = root << | ;
if(node[root].lazy) pushDown(root);
if(node[root].left == left && node[root].right == right) {
return node[root].sum;
}
if(left >= node[rson].left) {
return query(left, right, rson);
} else if(right <= node[lson].right) {
return query(left, right, lson);
} else {
return query(left, node[lson].right, lson) + query(node[rson].left, right, rson);
}
}
void clear(int n, const vector<int> &d) {
for(int i = ; i <= n; i ++) {
this->data[i] = d[i];
}
build(, n);
}
}; template <typename T>
class TreeToLink {
private:
struct Point {
int size, son, depth, father, top, newId;
T data;
} point[MAXN];
struct Edge {
int to, next;
} edge[MAXN << ];
int oldId[MAXN], first[MAXN], sign, sumOfPoint, cnt;
SegmentTree<T> st;
void dfs1(int u, int father = , int depth = ) {
point[u].depth = depth;
point[u].father = father;
point[u].size = ;
int maxson = -;
for(int i = first[u]; i != -; i = edge[i].next) {
int to = edge[i].to;
if(to == father) continue;
dfs1(to, u, depth + );
point[u].size += point[to].size;
if(point[to].size > maxson) {
point[u].son = to;
maxson = point[to].size;
}
}
}
void dfs2(int u, int top) {
point[u].newId = ++cnt;
oldId[cnt] = u;
point[u].top = top;
if(point[u].son == -) {
return ;
}
dfs2(point[u].son, top);
for(int i = first[u]; i != -; i = edge[i].next) {
int to = edge[i].to;
if(to == point[u].son || to == point[u].father) continue;
dfs2(to, to);
}
}
public:
void clear(int n) {
sumOfPoint = n;
sign = ;
cnt = ;
for(int i = ; i <= n; i ++) {
first[i] = -;
point[i].son = -;
scanf("%d", &point[i].data);
}
}
void addEdgeOneWay(int u, int v) {
edge[sign].to = v;
edge[sign].next = first[u];
first[u] = sign ++;
}
void addEdgeTwoWay(int u, int v) {
addEdgeOneWay(u, v);
addEdgeOneWay(v, u);
}
void preWork(int x = ) {
dfs1(x);
dfs2(x, x);
vector<int> data(sumOfPoint + );
for(int i = ; i <= sumOfPoint; i ++) {
data[i] = point[oldId[i]].data;
}
st.clear(sumOfPoint, data);
}
void updatePath(int x, int y, T z){
while(point[x].top != point[y].top){
if(point[point[x].top].depth < point[point[y].top].depth)
swap(x, y);
st.update(point[point[x].top].newId, point[x].newId, z);
x = point[point[x].top].father;
}
if(point[x].depth > point[y].depth)
swap(x, y);
st.update(point[x].newId, point[y].newId, z);
}
T queryPath(int x, int y){
T ans = ;
while(point[x].top != point[y].top){
if(point[point[x].top].depth < point[point[y].top].depth)
swap(x, y);
ans += st.query(point[point[x].top].newId, point[x].newId);
x = point[point[x].top].father;
}
if(point[x].depth > point[y].depth)
swap(x, y);
ans += st.query(point[x].newId, point[y].newId);
return ans;
}
void updateSon(int x, T z){
st.update(point[x].newId, point[x].newId + point[x].size - , z);
}
T querySon(int x){
return st.query(point[x].newId, point[x].newId + point[x].size - );
}
T queryPoint(int x) {
return queryPath(x, x);
}
void updatePoint(int x, T z) {
updatePath(x, x, z);
}
}; class Solution {
private:
int n, m, p;
TreeToLink<int> ttl;
public:
void solve() {
while(~scanf("%d%d%d", &n, &m, &p)){
ttl.clear(n);
for(int i = , a, b; i < m; i ++) {
scanf("%d%d", &a, &b);
ttl.addEdgeTwoWay(a, b);
}
ttl.preWork();
while(p --) {
char opt;
int a, b, c;
scanf(" %c%d", &opt, &a);
if(opt == 'I') {
scanf("%d%d", &b, &c);
ttl.updatePath(a, b, c);
} else if(opt == 'D') {
scanf("%d%d", &b, &c);
ttl.updatePath(a, b, -c);
} else if(opt == 'Q') {
printf("%d\n", ttl.queryPoint(a));
}
}
}
}
} DarkScoCu; int main() {
DarkScoCu.solve();
return ;
}
hdu 3966 Aragorn's Story : 树链剖分 O(nlogn)建树 O((logn)²)修改与查询的更多相关文章
- HDU 3966 Aragorn's Story 树链剖分+树状数组 或 树链剖分+线段树
HDU 3966 Aragorn's Story 先把树剖成链,然后用树状数组维护: 讲真,研究了好久,还是没明白 树状数组这样实现"区间更新+单点查询"的原理... 神奇... ...
- Hdu 3966 Aragorn's Story (树链剖分 + 线段树区间更新)
题目链接: Hdu 3966 Aragorn's Story 题目描述: 给出一个树,每个节点都有一个权值,有三种操作: 1:( I, i, j, x ) 从i到j的路径上经过的节点全部都加上x: 2 ...
- HDU 3966 Aragorn's Story(树链剖分)(线段树区间修改)
Aragorn's Story Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- HDU 3966 Aragorn's Story 树链剖分+BIT区间修改/单点询问
Aragorn's Story Description Our protagonist is the handsome human prince Aragorn comes from The Lord ...
- HDU 3966 Aragorn's Story 树链剖分
Link: http://acm.hdu.edu.cn/showproblem.php?pid=3966 这题注意要手动扩栈. 这题我交g++无限RE,即使手动扩栈了,但交C++就过了. #pragm ...
- hdu 3966 Aragorn's Story 树链剖分 按点
Aragorn's Story Time Limit: 10000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- HDU 3966 Aragorn's Story (树链剖分入门题)
树上路径区间更新,单点查询. 线段树和树状数组都可以用于本题的维护. 线段树: #include<cstdio> #include<iostream> #include< ...
- HDU 3966 Aragorn's Story (树链点权剖分,成段修改单点查询)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3966 树链剖分的模版,成段更新单点查询.熟悉线段树的成段更新的话就小case啦. //树链剖分 边权修 ...
- HDU 3966 Aragorn's Story 树链拋分
一.写在前面 终于开始开坑link-cut-tree这个了,对于网上找到的大佬的前进路线,进行了一番研发,发现实际上可以实现对于树链拋分的制作.经历了若干长时间之后终于打了出来(为什么每次学什么东西都 ...
随机推荐
- Servlet中listener(监听器)和filter的总结
Listener 我是这样理解他的,他是一种观察者模式的实现:我们在 web.xml 中配置 listener 的时候就是把一个被观察者放入的观察者的观察对象队列中,当被观察者触发了注册事件时观察者作 ...
- struts2返回结果类型
在action下还有result标签 1.result不只有name,其实还有type result返回类型在struts-default.xml默认的配置文件中有定义,可以看到有result-typ ...
- Error creating bean with name 'com.cloud.feign.interfaces.xxxFeignClient': FactoryBean threw exception on object creation; nested exception is java.lang.IllegalSt.PathVariable annotation was empty on
环境: Spring Cloud:Finchley.M8 Spring Boot:2.0.0.RELEASE 报错信息: Error creating bean with name 'com.clou ...
- git的commit规范及强制校验
1.背景 在多人协作项目中,如果代码风格统一.代码提交信息的说明准确,那么在后期协作以及Bug处理时会更加方便. 先来介绍本人公司采用的commit规范 Commit message格式 < ...
- Form表单元素
Form表单元素 action method input: name value type: text password button radio checkbox file submit reset ...
- 使用git版本管理工具
1.(1)正常提交:git add 提交文件 git init //git init之后建立一个.gitignore可以避免node_modules这类文件夹提交 git add . git co ...
- Linux 学习 一, useradd
安装好VMware 安装好Linux 在安装Linux时候,建立了一个用户,dragon, 和密码...这个用户不是root用户,没有root权限 可以切换dragon 到用户root,这个时候就有r ...
- MD5简单实例
如图当点击按钮时,会先判断是否第一次登陆,如果是第一次登陆登陆则会弹出设置密码的弹窗,若果登陆过则弹出登陆弹窗 其中输入的密码会用MD5加密下 package com.org.demo.wangfen ...
- 数据库系统异常排查之DMV(转)
来源: http://www.cnblogs.com/fygh/archive/2012/03/12.html 数据库系统异常是DBA经常要面临的情景,一名有一定从业经验的DBA,都会有自己一套故障排 ...
- Python 调度算法 死锁 静动态链接(七)
1 select poll epoll的区别 基本上select有3个缺点: 连接数受限 查找配对速度慢 数据由内核拷贝到用户态 poll改善了第一个缺点 epoll改了三个缺点. (1)select ...