Alyona and a tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alyona has a tree with n vertices. The root of the tree is the vertex 1. In each vertex Alyona wrote an positive integer, in the vertex i she wrote ai. Moreover, the girl wrote a positive integer to every edge of the tree (possibly, different integers on different edges).

Let's define dist(v, u) as the sum of the integers written on the edges of the simple path from v to u.

The vertex v controls the vertex u (v ≠ u) if and only if u is in the subtree of v and dist(v, u) ≤ au.

Alyona wants to settle in some vertex. In order to do this, she wants to know for each vertex v what is the number of vertices u such that v controls u.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105).

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the integers written in the vertices.

The next (n - 1) lines contain two integers each. The i-th of these lines contains integers pi and wi (1 ≤ pi ≤ n, 1 ≤ wi ≤ 109) — the parent of the (i + 1)-th vertex in the tree and the number written on the edge between pi and (i + 1).

It is guaranteed that the given graph is a tree.

Output

Print n integers — the i-th of these numbers should be equal to the number of vertices that the i-th vertex controls.

Examples
Input
5
2 5 1 4 6
1 7
1 1
3 5
3 6
Output
1 0 1 0 0
Input
5
9 7 8 6 5
1 1
2 1
3 1
4 1
Output
4 3 2 1 0
Note

In the example test case the vertex 1 controls the vertex 3, the vertex 3 controls the vertex 5 (note that is doesn't mean the vertex 1 controls the vertex 5).

【分析】题意就不说了,说下怎么做。很显然暴力肯定超时。对于每一个节点,用倍增的思想,fa[i][x]、cost[i][x]数组分别记录x节点往上第2的i次方个祖先编号和与那个祖先的距离。如果节点x对节点f有一个贡献,那么x对处于x和f之间的节点也应该有一个贡献,所以用前缀数组的思想,ans[x]++,ans[f]--,然后只需要在dfs的时候一个一个累加就行了。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 2e5+;
int n,m,k,u;
int fa[][N];
ll cost[][N];
ll a[N],ans[N];
struct man{
int son;
ll co;
};
vector<man>edg[N];
void dfs(int x){
for(int i=;fa[i-][fa[i-][x]];i++){
fa[i][x]=fa[i-][fa[i-][x]];
cost[i][x]=cost[i-][x]+cost[i-][fa[i-][x]];
}
for(int i=;i<edg[x].size();i++){
man e=edg[x][i];
int v=e.son;ll c=e.co;
fa[][v]=x;cost[][v]=c;
dfs(v);
}
}
int Find(int x){
ll c=a[x];
for(int i=;i>=;i--){
if(cost[i][x]<=c&&fa[i][x]){
c-=cost[i][x];
x=fa[i][x];
}
}
return x;
}
void sum_dfs(int x){
for(int i=;i<edg[x].size();i++){
man e=edg[x][i];
int v=e.son;ll c=e.co;
sum_dfs(v);
ans[x]+=ans[v];
}
}
int main (){
ll c;
met(ans,);
scanf("%d",&n);
for(int i=;i<=n;i++)scanf("%lld",&a[i]);
for(int i=;i<=n;i++){
scanf("%d%lld",&u,&c);
man s;s.co=c;s.son=i;
edg[u].push_back(s);
}
dfs();
for(int i=;i<=n;i++){
int f=Find(i);
ans[fa[][f]]--;
ans[fa[][i]]++;
}
sum_dfs();
for(int i=;i<=n;i++)printf("%lld ",ans[i]);printf("\n");
return ;
}

codeforces 381 D Alyona and a tree(倍增)(前缀数组)的更多相关文章

  1. 【30.36%】【codeforces 740D】Alyona and a tree

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  2. codeforces 682C C. Alyona and the Tree(dfs)

    题目链接: C. Alyona and the Tree time limit per test 1 second memory limit per test 256 megabytes input ...

  3. 【CodeForces - 682C】Alyona and the Tree(dfs)

    Alyona and the Tree Descriptions 小灵决定节食,于是去森林里摘了些苹果.在那里,她意外地发现了一棵神奇的有根树,它的根在节点 1 上,每个节点和每条边上都有一个数字. ...

  4. 【Codeforces 682C】Alyona and the Tree

    [链接] 我是链接,点我呀:) [题意] 题意 [题解] 设dis[v]表示v以上的点到达这个点的最大权值(肯定是它的祖先中的某个点到这个点) 类似于最大连续累加和 当往下走(x,y)这条边的时候,设 ...

  5. Codeforces Round #381 (Div. 2) D. Alyona and a tree 树上二分+前缀和思想

    题目链接: http://codeforces.com/contest/740/problem/D D. Alyona and a tree time limit per test2 secondsm ...

  6. Codeforces Round #381 (Div. 1) B. Alyona and a tree dfs序 二分 前缀和

    B. Alyona and a tree 题目连接: http://codeforces.com/contest/739/problem/B Description Alyona has a tree ...

  7. XJOI 3363 树4/ Codeforces 739B Alyona and a tree(树上差分+路径倍增)

    D. Alyona and a tree time limit per test  2 seconds memory limit per test  256 megabytes input  stan ...

  8. Codeforces Round #381 (Div. 2)D. Alyona and a tree(树+二分+dfs)

    D. Alyona and a tree Problem Description: Alyona has a tree with n vertices. The root of the tree is ...

  9. Codeforces Round #381 (Div. 2) D. Alyona and a tree dfs序+树状数组

    D. Alyona and a tree time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

随机推荐

  1. Autofac 解释第一个例子 《第一篇》

    Autofac是一个轻量级的依赖注入的框架,同类型的框架还有Spring.NET,Unity,Castle等. Autofac的使用有一个非常让人郁闷的地方,就是服务器要求安装有Microsoft . ...

  2. 移动端web页面如何适配

    移动端web页面如何适配,现有两个方案: 1 设置viewport进行缩放 简单粗暴,使用过程中反应缩放会导致有些页面元素会糊的情况.天猫的web app的首页使用这种方案 在页面中加入viewpor ...

  3. Eclipse SVN 安装步骤

    1.在eclipse中安装svn插件 Eclipse ---> Help ---> Install New Software ---> Add Name : 任意 Location ...

  4. ASP.NET MVC Html.Partial/Html.RenderPartial/Html.Action/Html.RenderAction区别

    1. @Html.Raw() 方法输出带有html标签的字符串: <div style="margin:10px 0px 0px;border:1px;border-color:red ...

  5. SG函数 专题练习

    [hdu1536][poj2960]S-Nim 题意 题意就是给出一个数组h,为每次可以取石子的数目. 然后给你n堆石子每堆si.求解先手能不能赢? 分析 根据\(h\)数组预处理出\(sg[i]\) ...

  6. Yii 提示Invalid argument supplied for foreach() 等错误

    Yii 提示Invalid argument supplied for foreach() 或者 undefined variable: val等错误 只需要在对应的文件中加入error_report ...

  7. Redis + php扩展的安装与配置(windows)

    -->安装Redis服务 下载redis安装包 http://pan.baidu.com/s/1pJiVFHx 下载后解压 把解压后文件夹里面的文件(根据自己的系统位数选择32位或者64位)拷贝 ...

  8. apche的主配置文件)

    apche的主配置文件conf/httpd.conf(根据个人主机的路径设置,以下仅供参考) 需配置的行号与方法(示列): 172  #ServerName localhost:80 173 Serv ...

  9. wpf中UserControl的几种绑定方式

    我们经常会抽取一些可重用的控件,某个属性是否需要重用,直接决定了这个属性的绑定方式. 1.完全不可重用的控件 有一些与业务强相关的控件,它们的属性完全来自ViewModel,越是相对复杂的控件,越容易 ...

  10. DeepLearning之路(二)SoftMax回归

    Softmax回归   1. softmax回归模型 softmax回归模型是logistic回归模型在多分类问题上的扩展(logistic回归解决的是二分类问题). 对于训练集,有. 对于给定的测试 ...