@(XSY)[分塊, 倍增]

Description

There's a new trend among Bytelandian schools. The "Byteland Touristic Bureau" has developed a new project for the high-schoolers. The project is so-called "Children's Trips".

The project itself is very simple: there are some touristic routes in Byteland, and N touristic campuses (numbered from 1 to N). For the sake of economy, there are exactly N-1 road between them. Of course, even having this given, it is possible to travel from any touristic campus to any other one. Moreover, for the sake of safety, each road is no longer than 2 kilometers.

When some student wants to travel, he first chooses his starting campus - he is been delivered there (say) by a helicopter. He chooses his final destination campus as well. From his final destination, he will be transported to his home by (say) a helicopter, again. So that pupil won't travel any extra distance by foot. When the start and the finish are chosen and the pupil is delivered, he starts his moving by the only route. None of pupils is infinitely strong, so first the pupil looks at the map of the touristic routes, and then he chooses the furthest campus on his way that he can reach during the current day (by safety regulations, it is strictly prohibited to sleep not at the campus because there can be a little trouble with werewolves), and moves there. Then the new day begins, and it repeats till the moment when the destination is reached.

Of course, not all the students created equal. Somebody is good in math, somebody in English, somebody in PE. So it's quite natural that all high-schoolers has different strengths.

We call the strength is the maximal number of kilometers that the pupil can pass in a day. And now you're given a lot of queries from the children. For every query, you are given the starting campus, the final campus and the strength. You are requested to calculate the number of days for every trip. The map of the campuses and the distances between them will be given to you as well.

Input

The first line of input contains the integer \(N\), denoting the number of campuses.

The next \(N-1\) lines contain triples of the form \(X\) \(Y\) \(D\) with the meaning that there is a road between the X-th and the Y-th campus with the length \(D\) kilometers.

Then there is a line with a single integer \(M\), denoting the number of queries.

Then, there are \(M\) lines with the triples of the form \(S\) \(F\) \(P\) with the meaning that the trip starts at the campus \(S\), ends at the campus \(F\) and the student has the strength of \(P\).

\[Output
For every query, please output the number of days on a separate line.
Constraints
$1 ≤ N, M ≤ 100000$
$1 ≤ X, Y, S, F ≤ N$
$1 ≤ D ≤ 2$
$2 ≤ P ≤ 2*N$
##Example
###Input:
```dos
5
1 2 1
2 3 2
1 4 2
4 5 1
5
1 5 3
1 3 2
2 5 4
1 2 10
4 5 2
```
###Output:
```dos
1
2
1
1
1
```
##Solution
翻譯一下題意:
>有一棵每条边的边权分别为$1$或$2$的共有$n$个节点的树, 对于一个询问, 给出起点$u$和终点$v$, 以及每一天最多走的路程$k$. 规定每天的结束点必须在树的节点上, 问最少要几天走完所有路程.

由於邊的權值只能為$1$或$2$, 因此可以考慮採用時間複雜度帶根號的算法.
>对$p$分情况进行讨论
>1. $p \ge \sqrt{n}$
>最多总共走$\sqrt{n}$天, 对于每一天倍增最远可以走到哪里即可, 并统计走的天数.
>2. $p \le sqrt(n)$
>对于每次询问的$p$, 暴力求出从每个点出发, 最多走$p$距离最远可以到达的最远位置. 再用倍增求出$f[i][j]$数组, 表示从点$i$出发, 走$2 ^ j$天可以到达的最远位置. 最后跑一次倍增解决.
>注意在进行这种计算之前, 先要将$p$从小到大排序. 这样每次计算就可以省去之前已经进行过的部分. 同时, 对于两个相同的$p$值, 不要重复求$f$数组, 否则会超时.

>时间复杂度$O \left( n * \sqrt{n} * log(n) \right)$

```cpp
#include<cstdio>
#include<cctype>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

inline int read()
{
int x = 0, flag = 1;
char c;
while(! isdigit(c = getchar()))
if(c == '-')
flag *= - 1;
while(isdigit(c))
x = x * 10 + c - '0', c = getchar();
return x * flag;
}

void println(int x)
{
if(x < 0)
putchar('-'), x *= - 1;
if(x == 0)
putchar('0');
int ans[1 << 5], top = 0;
while(x)
ans[top ++] = x % 10, x /= 10;
for(; top; top --)
putchar(ans[top - 1] + '0');
putchar('\n');
}

const int N = 1 << 17, M = 1 << 17;

int n;

int head[N];
int top;

struct edge
{
int v, w, next;
}G[N << 1];

inline void addEdge(int u, int v, int w)
{
G[top].v = v, G[top].w = w, G[top].next = head[u];
head[u] = top ++;
}

struct data
{
int u, dis;
}st[N][17];

int dep[N];
int disToRoot[N];

void dfs(int u, int pre, int w)
{
st[u][0].u = pre;
st[u][0].dis = w;
dep[u] = dep[pre] + 1;
disToRoot[u] = disToRoot[pre] + w;

for(int i = head[u]; ~ i; i = G[i].next)
if(G[i].v != pre)
dfs(G[i].v, u, G[i].w);
}

inline void getSt()
{
for(int i = 1; i < 17; i ++)
for(int j = 1; j <= n; j ++)
st[j][i].u = st[st[j][i - 1].u][i - 1].u,
st[j][i].dis = st[j][i - 1].dis + st[st[j][i - 1].u][i - 1].dis;
}

struct query
{
int id, s, t, p;

inline friend int operator <(query a, query b)
{
return a.p < b.p;
}
}a[M];

int getLca(int u, int v)
{
if(dep[u] < dep[v])
swap(u, v);

for(int i = 17 - 1; ~ i; i --)
if(dep[u] - (1 << i) >= dep[v])
u = st[u][i].u;

if(u == v)
return u;

for(int i = 17 - 1; ~ i; i --)
if(st[u][i].u != st[v][i].u)
u = st[u][i].u, v = st[v][i].u;

return st[u][0].u;
}

inline int climb(int &u, int lca, int p)
{
int ret = 0;

while(1)
{
if(disToRoot[u] - disToRoot[lca] < p)
break;

int left = p;

for(int i = 17 - 1; ~ i; i --)
if(left >= st[u][i].dis)
left -= st[u][i].dis, u = st[u][i].u;

ret ++;
}

return ret;
}

int ans[M];

int f[N][17];

inline int jump(int &u, int lca)
{
int ret = 0;

for(int i = 17 - 1; ~ i; i --)
if(dep[f[u][i]] > dep[lca])
ret += 1 << i, u = f[u][i];

return ret;
}

int main()
{
#ifndef ONLINE_JUDGE
freopen("childrenTrip.in", "r", stdin);
freopen("childrenTrip.out", "w", stdout);
#endif

n = read();
memset(head, - 1, sizeof(head));
top = 0;

for(int i = 1; i < n; i ++)
{
int u = read(), v = read(), w = read();
addEdge(u, v, w), addEdge(v, u, w);
}

dep[1] = - 1;
disToRoot[1] = 0;
dfs(1, 1, 0);
getSt();

int m = read();

for(int i = 0; i < m; i ++)
a[i].id = i, a[i].s = read(), a[i].t = read(), a[i].p = read();

sort(a, a + m);
int p;

for(p = 0; p < m; p ++)
if(a[p].p > (int)sqrt(n))
break;

for(int i = p; i < m; i ++)
{
int lca = getLca(a[i].s, a[i].t);
ans[a[i].id] = climb(a[i].s, lca, a[i].p) + climb(a[i].t, lca, a[i].p)
+ (disToRoot[a[i].s] + disToRoot[a[i].t] - 2 * disToRoot[lca] + a[i].p - 1) / a[i].p;
}

for(int i = 1; i <= n; i ++)
f[i][0] = i;

int last = 0;

for(int i = 0; i < p; i ++)
{
if(a[i].p != last)
{
for(int j = 1; j <= n; j ++)
{
int rest = a[i].p - (disToRoot[j] - disToRoot[f[j][0]]);
int u = f[j][0];

for(; ;)
{
rest -= st[u][0].dis;
u = st[u][0].u;

if(rest >= 0)
f[j][0] = u;

if(rest <= 0 || ! dep[u])
break;
}
}

for(int j = 1; j < 17; j ++)
for(int k = 1; k <= n; k ++)
f[k][j] = f[f[k][j - 1]][j - 1];

last = a[i].p;
}

int lca = getLca(a[i].s, a[i].t);

ans[a[i].id] = jump(a[i].s, lca) + jump(a[i].t, lca)
+ (disToRoot[a[i].s] + disToRoot[a[i].t] - 2 * disToRoot[lca] + a[i].p - 1) / a[i].p;
}

for(int i = 0; i < m; i ++)
println(ans[i]);
}
```\]

CODECHEF Oct. Challenge 2014 Children Trips的更多相关文章

  1. codechef January Challenge 2014 Sereja and Graph

    题目链接:http://www.codechef.com/JAN14/problems/SEAGRP [题意] 给n个点,m条边的无向图,判断是否有一种删边方案使得每个点的度恰好为1. [分析] 从结 ...

  2. Codechef March Challenge 2014——The Street

    The Street Problem Code: STREETTA https://www.codechef.com/problems/STREETTA Submit Tweet All submis ...

  3. 【分块+树状数组】codechef November Challenge 2014 .Chef and Churu

    https://www.codechef.com/problems/FNCS [题意] [思路] 把n个函数分成√n块,预处理出每块中各个点(n个)被块中函数(√n个)覆盖的次数 查询时求前缀和,对于 ...

  4. CodeChef November Challenge 2014

    重点回忆下我觉得比较有意义的题目吧.水题就只贴代码了. Distinct Characters Subsequence 水. 代码: #include <cstdio> #include ...

  5. 刷漆(Codechef October Challenge 2014:Remy paints the fence)

    [问题描述] Czy做完了所有的回答出了所有的询问,结果是,他因为脑力消耗过大而变得更虚了:).帮助Czy恢复身材的艰巨任务落到了你的肩上. 正巧,你的花园里有一个由N块排成一条直线的木板组成的栅栏, ...

  6. [Codechef October Challenge 2014]刷漆

    问题描述 Czy做完了所有的回答出了所有的询问,结果是,他因为脑力消耗过大而变得更虚了:).帮助Czy恢复身材的艰巨任务落到了你的肩上. 正巧,你的花园里有一个由N块排成一条直线的木板组成的栅栏,木板 ...

  7. Codechef December Challenge 2014 Chef and Apple Trees 水题

    Chef and Apple Trees Chef loves to prepare delicious dishes. This time, Chef has decided to prepare ...

  8. AC日记——The Street codechef March challenge 2014

    The Street 思路: 动态开节点线段树: 等差序列求和于取大,是两个独立的子问题: 所以,建两颗线段树分开维护: 求和:等差数列的首项和公差直接相加即可: 取大: 对于线段树每个节点储存一条斜 ...

  9. CODECHEF Nov. Challenge 2014 Chef & Churu

    @(XSY)[分塊] Hint: 題目原文是英文的, 寫得很難看, 因此翻譯為中文. Input Format First Line is the size of the array i.e. \(N ...

随机推荐

  1. Java-basic-2-

    接口只定义派生要用到的方法,但是方法的具体实现完全取决于派生类. 如果一个类定义在某个包中,那么package语句应该在源文件的首行. 如果源文件包含import语句,那么应该放在package语句和 ...

  2. Oracle redo与undo 第一弹

      一. 什么是redo(用于前滚数据) redo也就是重做日志文件(redo log file),Oracle维护着两类重做日志文件:在线(online)重做日志文件和归档(archived)重做日 ...

  3. 【HIHOCODER 1182】欧拉路·三

    描述 小Hi和小Ho破解了一道又一道难题,终于来到了最后一关.只要打开眼前的宝箱就可以通关这个游戏了. 宝箱被一种奇怪的机关锁住: 这个机关是一个圆环,一共有2^N个区域,每个区域都可以改变颜色,在黑 ...

  4. 水题:CF16C-Monitor

    Monitor 题目描述 Reca company makes monitors, the most popular of their models is AB999 with the screen ...

  5. poj 2385 树上掉苹果问题 dp算法

    题意:有树1 树2 会掉苹果,奶牛去捡,只能移动w次,开始的时候在树1 问最多可以捡多少个苹果? 思路: dp[i][j]表示i分钟移动j次捡到苹果的最大值 实例分析 0,1  1,2...说明 偶数 ...

  6. debian软raid

    http://www.linuxidc.com/Linux/2013-06/86487.htm  

  7. 使用docker+tomcat部署jenkins

  8. BugBash活动分享

    此文已由作者夏君授权网易云社区发布. 欢迎访问网易云社区,了解更多网易技术产品运营经验. BugBash源至微软概念,翻译为<缺陷大扫除>,顾名思义是集中大家力量全面清扫Bug,确保产品质 ...

  9. Jquery+Ajax+asp.net+sqlserver-编写的通用邮件管理(有源码)

    开始 邮件管理通常用在各个内部系统中,为了方便快捷的使用现有的代码开发一个邮件管理系统而诞生的. 准备条件 这是我的设计表结构,大家一看就懂了 --邮件接收表CREATE TABLE [dbo].[T ...

  10. Leetcode 421.数组中两数的最大异或值

    数组中两数的最大异或值 给定一个非空数组,数组中元素为 a0, a1, a2, … , an-1,其中 0 ≤ ai < 231 . 找到 ai 和aj 最大的异或 (XOR) 运算结果,其中0 ...