Tourists

时间限制: 5 Sec  内存限制: 64 MB

题目描述

In Tree City, there are n tourist attractions uniquely labeled 1 to n. The attractions are connected by a set of n − 1 bidirectional roads in such a way that a tourist can get from any attraction to any other using some path of roads.
You are a member of the Tree City planning committee. After much research into tourism, your committee has discovered a very interesting fact about tourists: they LOVE number theory! A tourist who visits an attraction with label x will then visit another attraction with label y if y > x and y is a multiple of x. Moreover, if the two attractions are not directly connected by a road thetourist will necessarily visit all of the attractions on the path connecting x and y, even if they aren’t multiples of x. The number of attractions visited includes x and y themselves. Call this the length of a path.
Consider this city map:

Here are all the paths that tourists might take, with the lengths for each:
1 → 2 = 4, 1 → 3 = 3, 1 → 4 = 2, 1 → 5 = 2, 1 → 6 = 3, 1 → 7 = 4,
1 → 8 = 3, 1 → 9 = 3, 1 → 10 = 2, 2 → 4 = 5, 2 → 6 = 6, 2 → 8 = 2,
2 → 10 = 3, 3 → 6 = 3, 3 → 9 = 3, 4 → 8 = 4, 5 → 10 = 3
To take advantage of this phenomenon of tourist behavior, the committee would like to determine the number of attractions on paths from an attraction x to an attraction y such that y > x and y is a multiple of x. You are to compute the sum of the lengths of all such paths. For the example above, this is: 4 + 3 + 2 + 2 + 3 + 4 + 3 + 3 + 2 + 5 + 6 + 2 + 3 + 3 + 3 + 4 + 3 = 55.

输入

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The first line of input will consist of an integer n (2 ≤ n ≤ 200,000) indicating the number of attractions. Each of the following n−1 lines will consist of a pair of space-separated
integers i and j (1 ≤ i < j ≤ n), denoting that attraction i and attraction j are directly connected by a road. It is guaranteed that the set of attractions is connected.

输出

Output a single integer, which is the sum of the lengths of all paths between two attractions x and y such that y > x and y is a multiple of x.

样例输入

10
3 4
3 7
1 4
4 6
1 10
8 10
2 8
1 5
4 9

样例输出

55
分析:LCA裸题;
   注意dfs层数太深会爆,所以需要手写栈;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
#define sys system("pause")
const int maxn=2e5+;
using namespace std;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}
inline ll read()
{
ll x=;int f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,k,t,p[maxn<<],all,dep[maxn],tot,h[maxn],vis[maxn<<],vis1[maxn],fa[maxn],st[][maxn<<];
void init()
{
for(int i=;i<=all;i++)p[i]=+p[i>>];
for(int i=;i<=;i++)
for(int j=;(j+(<<i)-)<=(ll)all;j++)
st[i][j]=min(st[i-][j],st[i-][j+(<<(i-))]);
}
int query(int l,int r)
{
int x=p[r-l+];
return min(st[x][l],st[x][r-(<<x)+]);
}
struct node
{
int to,nxt;
}e[maxn<<];
void add(int x,int y)
{
tot++;
e[tot].to=y;
e[tot].nxt=h[x];
h[x]=tot;
}
stack<int>S;
void dfs()
{
S.push();
while(!S.empty())
{
int now = S.top();
if(vis1[now] == )// if node is gray, then color black
{
vis1[now] = ;
st[][++all]=dep[fa[now]];
// do things after dfs children.
S.pop();
}
else if(vis1[now] == )// if node is white, then color gray
{
vis1[now] = ;
st[][++all]=dep[now];
vis[now]=all;
// do things before dfs children.
for(int i=h[now];i;i=e[i].nxt)
{
int to=e[i].to;
if(!vis1[to])
{
dep[to]=dep[now]+;
fa[to]=now;
S.push(to);
}
}
}
}
}
ll ans;
int main()
{
int i,j;
//freopen("in.txt","r",stdin);
while(~scanf("%d",&n))
{
ans=;
all=;
tot=;
memset(dep,,sizeof(dep));
memset(vis,,sizeof(vis));
memset(h,,sizeof(h));
memset(p,,sizeof(p));
memset(vis1,,sizeof(vis1));
rep(i,,n-)
{
int a,b;
scanf("%d%d",&a,&b);
add(a,b),add(b,a);
}
dfs();
init();
for(i=;i<=n;i++)
{
for(j=i*;j<=n;j+=i)
{
ans+=dep[i]+dep[j]-*query(min(vis[i],vis[j]),max(vis[i],vis[j]))+;
}
}
printf("%lld\n",ans);
}
//system("Pause");
return ;
}

Tourists的更多相关文章

  1. 【Codefoces487E/UOJ#30】Tourists Tarjan 点双连通分量 + 树链剖分

    E. Tourists time limit per test: 2 seconds memory limit per test: 256 megabytes input: standard inpu ...

  2. CF487 E. Tourists [点双连通分量 树链剖分 割点]

    E. Tourists 题意: 无向连通图 C a w: 表示 a 城市的纪念品售价变成 w. A a b: 表示有一个游客要从 a 城市到 b 城市,你要回答在所有他的旅行路径中最低售价的最低可能值 ...

  3. 【CF487E】Tourists(圆方树)

    [CF487E]Tourists(圆方树) 题面 UOJ 题解 首先我们不考虑修改,再来想想这道题目. 我们既然要求的是最小值,那么,在经过一个点双的时候,走的一定是具有较小权值的那一侧. 所以说,我 ...

  4. 每日英语:Foreign Tourists Skip Beijing

    Overseas tourists continued to shun Beijing through 2013. shun:避开,避免,回避 Amid rising pollution and a ...

  5. L192 Virgin Galactic Completes Test of Spaceship to Carry Tourists

    Virgin Galactic says its spacecraft designed to launch tourists into space completed an important te ...

  6. UOJ #30. [CF Round #278] Tourists

    UOJ #30. [CF Round #278] Tourists 题目大意 : 有一张 \(n\) 个点, \(m\) 条边的无向图,每一个点有一个点权 \(a_i\) ,你需要支持两种操作,第一种 ...

  7. [UOJ30]/[CF487E]Tourists

    [UOJ30]/[CF487E]Tourists 题目大意: 一个\(n(n\le10^5)\)个点\(m(m\le10^5)\)条边的无向图,每个点有点权.\(q(q\le10^5)\)次操作,操作 ...

  8. Tourists——圆方树

    CF487E Tourists 一般图,带修求所有简单路径代价. 简单路径,不能经过同一个点两次,那么每个V-DCC出去就不能再回来了. 所以可以圆方树,然后方点维护一下V-DCC内的最小值. 那么, ...

  9. 【学习笔记】圆方树(CF487E Tourists)

    终于学了圆方树啦~\(≧▽≦)/~ 感谢y_immortal学长的博客和帮助 把他的博客挂在这里~ 点我传送到巨佬的博客QwQ! 首先我们来介绍一下圆方树能干什么呢qwq 1.将图上问题简化到树上问题 ...

随机推荐

  1. MPI编程简介[转]

    原文地址http://blog.csdn.net/qinggebuyao/article/details/8059300 3.1 MPI简介 多线程是一种便捷的模型,其中每个线程都可以访问其它线程的存 ...

  2. 解决IOS下不支持fixed的问题

    我们公司有一个页面底部用到了fixed样式,每当弹出键盘的时候,IOS下fixed就会走样(据我所知android没有该问题). 为此之前我经过产品的同意做了简单的处理(方法1). 方法一: focu ...

  3. javascript apply()和call()

    原文链接 http://www.jb51.net/article/30883.htm 想要理解透彻apply()和call() ,还要需要理解this  作用域 局部变量  全局变量 js apply ...

  4. mybatis 批量插入值的sql

    <insert id="insertAwardPic" useGeneratedKeys="true" parameterType="java. ...

  5. 浙江大学 pat 1006题解

    1006. Sign In and Sign Out (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue ...

  6. 我使用的Bash脚本模板

    在Linux环境下(包括在相应的模拟环境,如Windows中的Cygwin)工作的时间久了,就会慢慢积累出一些自己写的小脚本程序,用于加速日常的操作与工作流程,如自动挂载与卸载U盘,映射网络驱动器,数 ...

  7. Gentoo/Funtoo USE标记介绍

    Gentoo/Funtoo USE标记 USE的简单理解如下:一个软件不只包含软件本身,还包括其组件,如,文档,插件,GUI支持等.USE就是用来标记是否要安装软件的同时安装这些组件. 声明USE标记 ...

  8. Hanoi汉诺塔问题——递归与函数自调用算法

    题目描述 Description 有N个圆盘,依半径大小(半径都不同),自下而上套在A柱上,每次只允许移动最上面一个盘子到另外的柱子上去(除A柱外,还有B柱和C柱,开始时这两个柱子上无盘子),但绝不允 ...

  9. sql身份证号查人数

    select a.*,b.item_name from (select sum(buy_num) num,nqh from (select substr(a.embed_dis_province,1, ...

  10. 十八、oracle 角色

    一.介绍角色就是相关权限的命令集合,使用角色的主要目的就是为了简化权限的管理.假定有用户a,b,c为了让他们都拥有如下权限1. 连接数据库2. 在scott.emp表上select,insert,up ...