E - Vasya and a Tree

思路:

dfs动态维护关于深度树状数组

返回时将当前节点的所有操作删除就能保证每次访问这个节点时只进行过根节点到当前节点这条路径上的操作

代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head const int N = 3e5 + ;
vector<int> g[N];
vector<pii> que[N];
LL bit[N], ans[N];
int n;
void add(int x, int d) {
while(x <= n) bit[x] += d, x += x&-x;
}
LL sum(int x) {
LL ans = ;
while(x) ans += bit[x], x -= x&-x;
return ans;
} void dfs(int u, int o, int deep) {
for (pii t : que[u]) {
int d = t.fi;
int x = t.se;
int l = deep;
int r = min(deep+d, n);
add(l, x);
add(r+, -x);
}
ans[u] = sum(deep);
for (int v : g[u]) {
if(v != o) dfs(v, u, deep+);
}
for (pii t : que[u]) {
int d = t.fi;
int x = t.se;
int l = deep;
int r = min(deep+d, n);
add(l, -x);
add(r+, +x);
}
}
int main() {
int u, m, v, d, x;
scanf("%d", &n);
for (int i = ; i < n; i++) {
scanf("%d %d", &u, &v);
g[u].pb(v);
g[v].pb(u);
}
scanf("%d", &m);
for (int i = ; i <= m; i++) {
scanf("%d %d %d", &v, &d, &x);
que[v].pb({d, x});
}
dfs(, , );
for (int i = ; i <= n; i++) printf("%lld%c", ans[i], " \n"[i==n]);
return ;
}

Codeforces 1076 E - Vasya and a Tree的更多相关文章

  1. Vasya and a Tree CodeForces - 1076E(线段树+dfs)

    I - Vasya and a Tree CodeForces - 1076E 其实参考完别人的思路,写完程序交上去,还是没理解啥意思..昨晚再仔细想了想.终于弄明白了(有可能不对 题意是有一棵树n个 ...

  2. Vasya and a Tree CodeForces - 1076E (线段树 + dfs)

    题面 Vasya has a tree consisting of n vertices with root in vertex 1. At first all vertices has 0 writ ...

  3. CodeForces-1076E Vasya and a Tree

    CodeForces - 1076E Problem Description: Vasya has a tree consisting of n vertices with root in verte ...

  4. codeforces 1065F Up and Down the Tree

    题目链接:codeforces 1065F Up and Down the Tree 题意:给出一棵树的节点数\(n\)以及一次移动的最大距离\(k\),现在有一个标记在根节点1处,每一次可以进行一下 ...

  5. Codeforces1076E. Vasya and a Tree(dfs+离线+动态维护前缀和)

    题目链接:传送门 题目: E. Vasya and a Tree time limit per test seconds memory limit per test megabytes input s ...

  6. CF1076E:Vasya and a Tree(DFS&差分)

    Vasya has a tree consisting of n n vertices with root in vertex 1 1 . At first all vertices has 0 0 ...

  7. Codeforces 914H Ember and Storm's Tree Game 【DP】*

    Codeforces 914H Ember and Storm's Tree Game 题目链接 ORZ佬 果然出了一套自闭题 这题让你算出第一个人有必胜策略的方案数 然后我们就发现必胜的条件就是树上 ...

  8. CF Edu54 E. Vasya and a Tree DFS+树状数组

    Vasya and a Tree 题意: 给定一棵树,对树有3e5的操作,每次操作为,把树上某个节点的不超过d的子节点都加上值x; 思路: 多开一个vector记录每个点上的操作.dfs这颗树,同时以 ...

  9. Codeforces Round #499 (Div. 1) F. Tree

    Codeforces Round #499 (Div. 1) F. Tree 题目链接 \(\rm CodeForces\):https://codeforces.com/contest/1010/p ...

随机推荐

  1. Golang对文件读写操作

    package main import ( "bufio" "fmt" "io" "os" ) //写 func Wri ...

  2. 01:jQuery的下拉选select2插件用法

    1.1 select2插件基本使用 1.下载select2插件 1. 下载地址:https://github.com/select2/select2 2.官网地址:https://select2.or ...

  3. php7安装redis拓展

    phpredis下载地址https://github.com/phpredis/phpredis   解压并进入源码包 unzip phpredis-develop.zip cd phpredis-d ...

  4. 线程同步——用户模式下线程同步——Interlocked实现线程同步

    线程同步分为用户模式下的线程同步和内核对象的线程同步. 当然用户模式下的线程同步实现速度比内核模式下快,但是功能也有局 //1.利用原子访问: Interlocked系列函数,关于Interlocke ...

  5. poj 1456 Supermarket - 并查集 - 贪心

    题目传送门 传送点I 传送点II 题目大意 有$n$个商品可以销售.每个商品销售会获得一个利润,但也有一个时间限制.每个商品需要1天的时间销售,一天也只能销售一件商品.问最大获利. 考虑将出售每个物品 ...

  6. 我的QML

    1.键盘加Text import QtQuick 2.7 import QtGraphicalEffects 1.0 Rectangle{ width:; height:; color:"# ...

  7. CentOS7下Docker中构建Jenkins容器

    背景 在CentOS搭建Docker完成后,我们需要在Docker中搭建Jenkins用来实现工程自动部署. 安装前提 jdk已安装,安装目录如:usr/java/jdk1.8.0_161 maven ...

  8. Bootstrap3基础 nav 便签页(横版、竖版)

      内容 参数   OS   Windows 10 x64   browser   Firefox 65.0.2   framework     Bootstrap 3.3.7   editor    ...

  9. Python3 tkinter基础 Frame event.x 输出鼠标左右键点击的位置

             Python : 3.7.0          OS : Ubuntu 18.04.1 LTS         IDE : PyCharm 2018.2.4       Conda ...

  10. 【学习】Hall’s Marriage Theorem

    其实是在做题时遇到这个定理的. 这个定理的图论意义是: 对于一个二分图\(G=\{X+Y,E\}\),它满足: \(\forall W \subseteq X, \, |W| \leq |N_G(W) ...