题目链接 : https://nanti.jisuanke.com/t/29228 There is an apple tree in Teemo's yard. It contains n nodes and n-1 branches, and the node 1 is always the root of the tree. Today, Teemo's father will go out for work. So Teemo should do his father's job in…

A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. An…

A Simple Tree Problem Time Limit: 3000ms Memory Limit: 65536KB This problem will be judged on ZJU. Original ID: 368664-bit integer IO format: %lld      Java class name: Main Prev Submit Status Statistics Discuss Next Type: None   None Graph Theory 2-…

I. A Simple Tree Problem Time Limit: 3000ms Memory Limit: 65536KB 64-bit integer IO format: %lld      Java class name: Main   Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of…

原题: ZOJ 3686 http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3686 这题本来是一个比较水的线段树,结果一个mark坑了我好几个小时..哎.太弱. 先DFS这棵树,树形结构转换为线性结构,每个节点有一个第一次遍历的时间和最后一次遍历的时间,之间的时间戳都为子树的时间戳,用线段树更新这段区间即可实现更新子树的效果,用到懒操作节省时间. 坑我的地方: update时,不能写成:tree[rt].mark = 1,…

A unival tree (which stands for "universal value") is a tree where all nodes under it have the same value. Given the root to a binary tree, count the number of unival subtrees. For example, the following tree has 5 unival subtrees: 0 / \ 1 0 / \…

Description Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. And we want to query the numbers of 1's of a subtree. Input Mu…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3686 题意:给定一颗有根树,每个节点有0和1两种值.有两种操作:o a操作,把以a为根节点的子树的权值全部取反:q a操作,求以a为根节点的子树权值为1的节点个数. 先求出树的先序遍历结果,并且记录每颗子树的节点个数,然后就可以用线段树维护了.. //STATUS:C++_AC_240MS_6524KB #include <functional> #inclu…

Solution: 根据树的遍历道的时间给树的节点编号,记录下进入节点和退出节点的时间.这个时间区间覆盖了这个节点的所有子树,可以当做连续的区间利用线段树进行操作. /* 线段树 */ #pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…

已经连咕了好几天博客了:比较经典的题目 题目大意 给出一个 N 个点的树和$K_i$, 求每个点到其他所有点距离中第 $K_i$ 小的数值. 题目分析 做法一:点分树上$\log^3$ 首先暴力做法:对于每个节点维护其他点距离的值域线段树.这个做法的瓶颈在于关于边$(u,v)$线段树的转移.那么可以利用点分树(为了保证复杂度)换一种容斥的思路利用重复的信息:记$f_i$为以$i$为根的点分树内所有其他点到点$i$的距离的值域线段树:$g_i$为以$i$为根的点分树内,所有点到$i$的点分树父亲的…