转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud

GCD Tree

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 84    Accepted Submission(s): 38

Problem Description
Teacher Mai has a graph with n vertices numbered from 1 to n. For every edge(u,v), the weight is gcd(u,v). (gcd(u,v) means the greatest common divisor of number u and v).

You need to find a subset of the edges that forms a tree that includes every vertex, where the total weight of all the edges in the tree is maximized. Print the total weight of these edges.

 



Input
There are multiple test cases(about 105).

For each test case, there is only one line contains one number n(1≤n≤105).

 



Output
For each test case, print the answer.
 



Sample Input
1
2
3
4
5
 



Sample Output
0
1
2
4
5

LCT维护最大生成树,从小到大不断加入数的时候,先和它最大的约数连边,这样就已经变成一棵树了,然后从大到小依次查询其他因子,每次查找到这个数的路径上的最小值,然后砍断这条边,把i和这次查询的因子相连,因为砍掉的边的权值一定不会超过新加的边,所以每次都在增大,这样能够保证最优

队友随手过了这题。23333333

我的写法是把边权搞成一个点,其实不这样也行,找出最小的点,然后看一下在路径上和这个点相邻的点。

 /**
* code generated by JHelper
* More info: https://github.com/AlexeyDmitriev/JHelper
* @author xyiyy @https://github.com/xyiyy
*/ #include <iostream>
#include <fstream> //#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype> using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define mp(X, Y) make_pair(X,Y)
#define pb(X) push_back(X)
#define rep(X, N) for(int X=0;X<N;X++)
#define rep2(X, L, R) for(int X=L;X<=R;X++)
#define dep(X, R, L) for(int X=R;X>=L;X--)
#define clr(A, X) memset(A,X,sizeof(A))
#define IT iterator
#define ALL(X) (X).begin(),(X).end()
#define PQ std::priority_queue
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<PII> VII;
typedef vector<int> VI; const int MAXN = ;
int pre[MAXN<<], ch[MAXN<<][], rev[MAXN<<];
int key[MAXN<<];
int lx[MAXN<<],rx[MAXN<<];
int mx[MAXN<<]; void push_down(int r) {
if(!r)return;
if (rev[r]) {
rev[ch[r][]] ^= ;
rev[ch[r][]] ^= ;
swap(ch[r][], ch[r][]);
rev[r] ^= ;
}
} void push_up(int x) {
int l = ch[x][],r = ch[x][];
mx[x] = x;
if(key[mx[l]]<key[mx[x]])mx[x] = mx[l];
if(key[mx[r]]<key[mx[x]])mx[x] = mx[r];
} void rotate(int x, int d) {
const int y = pre[x];
ch[y][!d] = ch[x][d];
if (ch[x][d])pre[ch[x][d]] = y;
pre[x] = pre[y];
if (ch[pre[y]][] == y)ch[pre[x]][] = x;
else if (ch[pre[y]][] == y)ch[pre[x]][] = x;
pre[y] = x;
ch[x][d] = y;
push_up(y);
} bool _splay_parent(int x,int &y){
return (y = pre[x])!= && (ch[y][] == x || ch[y][] == x);
} void splay(int x,int goal) {
push_down(x);
for (int y,z;_splay_parent(x,y);) {
//cout<<x<<" "<<y<<endl;
if(_splay_parent(y,z)){
push_down(z);push_down(y);push_down(x);
int d = y == ch[z][];
if(x == ch[y][d])rotate(x,d^),rotate(x,d);
else rotate(y,d),rotate(x,d);
}else {
push_down(y),push_down(x);
rotate(x, x == ch[y][]);break;
}
}
push_up(x);
} int access(int u) {
int v = ;
for (; u; u = pre[u]) {
splay(u,);
ch[u][] = v;
push_up(v = u);
}
return v;
} void makeroot(int x) {
rev[access(x)] ^= ;
splay(x,);
} void link(int x, int y) {
makeroot(x);
pre[x] = y;
} void cut(int x, int y) {
makeroot(x);
access(y);
splay(y,);
pre[ch[y][]] = ;
ch[y][] = ;
push_up(y);
} void Init(int n) {
for (int i = ; i < n; i++) {
pre[i] = ch[i][] = ch[i][] = ;
key[i] = INF;
mx[i] = ;
}
}
void debug(int x){ }
int query(int x, int y) {
makeroot(x);
access(y);
splay(y,);
return mx[y];
} vector<int> vec[MAXN];
int ans[MAXN]; class hdu5398 {
public:
void solve(std::istream &in, std::ostream &out) {
rep2(i, , MAXN-) {
vec[i].pb();
for (int j = ; j * j <= i; j++) {
if (i % j == ) {
vec[i].pb(j);
if (i / j != j)vec[i].pb(i / j);
}
}
sort(vec[i].begin(),vec[i].end());
}
Init(MAXN<<);
ans[] = ;
rep2(i, , MAXN-) {
int sz = vec[i].size();
int y = vec[i][sz - ];
ans[i] = ans[i - ];
link(i, MAXN + i);
rx[MAXN + i] = i;
link(y, MAXN + i);
lx[MAXN + i] = y;
key[MAXN + i] = y;
ans[i] += y;
dep(j, sz - , ) {
y = vec[i][j];
int x = query(y, i);
cut(x,lx[x]);
cut(x,rx[x]);
link(y, x);
link(i, x);
ans[i] -= key[x];
key[x] = y;
lx[x] = y;
rx[x] = i;
ans[i] += y;
}
}
int n;
while(in>>n){
out<<ans[n]<<endl;
} }
}; int main() {
std::ios::sync_with_stdio(false);
std::cin.tie();
hdu5398 solver;
std::istream &in(std::cin);
std::ostream &out(std::cout);
solver.solve(in, out);
return ;
}

hdu5398 GCD Tree(lct)的更多相关文章

  1. HDU 5002 Tree LCT 区间更新

    Tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?c ...

  2. BZOJ 3282: Tree( LCT )

    LCT.. -------------------------------------------------------------------------------- #include<c ...

  3. BZOJ 2631: tree( LCT )

    LCT...略麻烦... -------------------------------------------------------------------------------- #inclu ...

  4. [bzoj3282]Tree (lct)

    昨天看了一天的lct..当然幸好最后看懂了(也许吧..) 论善良学长的重要性T_T,老司机带带我! 这题主要是删边的时候还要判断一下..蒟蒻一开始天真的以为存在的边才能删结果吃了一发wa... 事实是 ...

  5. Link-Cut Tree(LCT)&TopTree讲解

    前言: Link-Cut Tree简称LCT是解决动态树问题的一种数据结构,可以说是我见过功能最强大的一种树上数据结构了.在此与大家分享一下LCT的学习笔记.提示:前置知识点需要树链剖分和splay. ...

  6. 动态树Link-cut tree(LCT)总结

    动态树是个好玩的东西 LCT题集 预备知识 Splay 树链剖分(好像关系并不大) 动态树(Link-cut tree) 先搬dalao博客 什么是LCT? 动态树是一类要求维护森林的连通性的题的总称 ...

  7. 洛谷P3690 [模板] Link Cut Tree [LCT]

    题目传送门 Link Cut Tree 题目背景 动态树 题目描述 给定n个点以及每个点的权值,要你处理接下来的m个操作.操作有4种.操作从0到3编号.点从1到n编号. 0:后接两个整数(x,y),代 ...

  8. 浅谈Link-Cut Tree(LCT)

    0XFF 前言&概念 Link-Cut Tree 是一种用来维护动态森林连通性的数据结构,适用于动态树问题.它采用类似树链剖分的轻重边路径剖分,把树边分为实边和虚边,并用 Splay 来维护每 ...

  9. Link-Cut Tree(LCT)

    转载自LCT(Link-Cut Tree)详解(蒟蒻自留地) 如果你还没有接触过LCT,你可以先看一看这里: (看不懂没关系,先留个大概的印像)http://www.cnblogs.com/BLADE ...

随机推荐

  1. str_repeat() 函数

    <?php echo str_repeat(".",13);//重复几次 ?>

  2. js 解析 json

    1.简单的json格式 { "user": [ { "name":"name1", "age":24, "se ...

  3. UIImagePickerController 操作图库

     UIImagePickerController详解   转载自:http://blog.csdn.net/kingsley_cxz/article/details/9157093 1.UIImage ...

  4. SVM及其对偶

    引自 http://my.oschina.net/wangguolongnk/blog/111349 1. 支持向量机的目的是什么? 对于用于分类的支持向量机来说,给定一个包含正例和反例(正样本点和负 ...

  5. 健身计划_from85to75

    第一天没什么好写的,这半年也没看什么书,就写写未来的规划好了. 当然是从最简单的健身计划开始写咯. 关键词:弹性 目标:减肥,上肢力量 时间:3-4次/周(Thur,Fri,Sat,Sun),1h-1 ...

  6. js 比较日期大小

    //1获取当前时间 var curTime = new Date(); //2把字符串格式转换为日期类 var startTime = new Date(Date.parse(kc.begintime ...

  7. JSP标准库标签 ———C标签

    一.C标签    一] <c:out value="..." default="..." escapeXml="true">   ...

  8. 在Activity的生命周期中,会被系统回调的方法

    onCreate(Bundle savedStatus):创建Activity时被回调.onStart():启动Activity时被回调.onRestart():重新启动Activity时被回调.on ...

  9. 颠覆你的时空观-----理解傅立叶transform

    在知乎上看到的,非常不错,最起码知道为什么了: 傅立叶变换,拉普拉斯变换,z变换这三种方法的本质主要就是将信号从时域转换成频域,因为频域更好展开分析 = =. 频域只是另一种看到信号的角度(世界观). ...

  10. Redis分布式缓存 教程以及DEMO

    原文地址:http://blog.csdn.net/qiujialongjjj/article/category/1800171 redis demo源码下载:http://download.csdn ...