The xor-longest Path Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10038   Accepted: 2040 Description In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p: ⊕ is the xor operator. W…

做该题之前,至少要先会做这道题. 记 \(d[u]\) 表示 \(1\) 到 \(u\) 简单路径的异或和,该数组可以通过一次遍历求得. \(~\) 考虑 \(u\) 到 \(v\) 简单路径的异或和该怎么求? 令 \(z=\operatorname{lca}(u,v)\) ,则 \(u\) 到 \(v\) 简单路径的异或和可以分成两段求解:一段是 \(z\) 到 \(u\) 简单路径的异或和,一段是 \(z\) 到 \(v\) 简单路径的异或和,二者异或一下即为 \(u\) 到 \(v\) 简…

We know that the longest path problem for general case belongs to the NP-hard category, so there is no known polynomial time solution for it. However, for a special case which is directed acyclic graph, we can solve this problem in linear time. First…

We all know that the shortest path problem has optimal substructure. The reasoning is like below: Supppose we have a path p from node u to v, another node t lies on path p: u->t->v ("->" means a path). We claim that u->t is also a sh…

题意 In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p: \[ _{xor}length(p)=\oplus_{e \in p}w(e) \] \(⊕\) is the xor operator. We say a path the xor-longest path if it has the largest xor-length.…

题解: 在树上i到j的异或和可以直接转化为i到根的异或和^j到根的异或和. 所以我们把每个点到根的异或和处理出来放到trie里面,再把每个点放进去跑一遍即可. 代码: #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<algorithm> #include<iostream> #include<vector> #inc…

给的多叉树, 找这颗树里面最长的路径长度 解法就是在子树里面找最大的两个(或一个,如果只有一个子树的话)高度加起来. 对于每一个treenode, 维护它的最高的高度和第二高的高度,经过该点的最大路径就是:  最高高度+第二高高度,然后return 最高高度 package fbPractise; import java.util.*; class TreeNode { int val; List<TreeNode> children; public TreeNode(int value) {…

应该是模板题了吧 定义: 树的直径是指一棵树上相距最远的两个点之间的距离. 方法:我使用的是比较常见的方法:两边dfs,第一遍从任意一个节点开始找出最远的节点x,第二遍从x开始做dfs找到最远节点的距离即为树的直径. 证明:假设此树的最长路径是从s到t,我们选择的点为u.反证法:假设第一遍搜到的点是v. 1.v在这条最长路径上,那么dis[u,v]>dis[u,v]+dis[v,s],显然矛盾. 2.v不在这条最长路径上,我们在最长路径上选择一个点为po,则dis[u,v]>dis[u,po]…

题意:给你一张DAG,求图中的最长路径. 题解:用拓扑排序一个点一个点的拿掉,然后dp记录步数即可. 代码: int n,m; int a,b; vector<int> v[N]; int in[N]; int dp[N]; int main() { //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); n=read(); m=read(); for(int i=1;i<=m;++i){ a=read(); b=read(); v[a…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

［抄题］: Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between th…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

题目: Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. The length of path between two nodes is represented by the number of edges between them. Exam…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 日期 题目地址:https://leetcode.com/problems/longest-univalue-path/description/ 题目描述 Given a binary tree, find the length of the longest path where each node in the path has the s…

os.path - Common pathname manipulations操作 This module implements some useful functions on pathnames. To read or write files see open(), and for accessing the filesystem see the os module. The path parameters can be passed as either strings, or bytes.…

题意:给出源点和边,边权为1,让你求从源点出发的最长路径,求出路径长度和最后地点,若有多组,输出具有最小编号的最后地点. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> using namespace std; ; int n,s,tot; //n:点的个数,s:源点 int dist[maxn]; /…

Ba Gua Zhen Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description During the Three-Kingdom period, there was a general named Xun Lu who belonged to Kingdom Wu. Once his troop were chasing Bei Liu, he was stuck in the Ba Gua Zhen from Liang Zhuge.…

  Longest Paths  It is a well known fact that some people do not have their social abilities completely enabled. One example is the lack of talent for calculating distances and intervals of time. This causes some people to always choose the longest w…

This module implements some useful functions on pathnames. To read or write files see open(), and for accessing the filesystem see the os module. The path parameters can be passed as either strings, or bytes. Applications are encouraged to represent…

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root. Note: The length of path between two nodes is represented by the number of edges between them. Ex…

Single Number 题目 Given an array of integers, every element appears twice except for one. Find that single one.      Note:      Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 思路: 为了满足时间和空间复杂度…

今天做了三道LeetCode上的简单题目,每道题都是用c++和Python两种语言写的.由于c++版的代码网上比較多.所以就仅仅分享一下Python的代码吧,刚学完Python的基本的语法,做做LeetCode的题目还是不错的,对以后找工作面试也有帮助! 刚開始就从AC率最高的入手吧! 1.Given an array of integers, every element appears twice except for one. Find that single one. Note: Your…

转载请注明原文地址:http://www.cnblogs.com/ygj0930/p/6412505.html 心得:看到一道题,优先往栈,队列,map,list这些工具的使用上面想.不要来去都是暴搜,数组遍历. 11:Single Number Given an array of integers, every element appears twice except for one. Find that single one. 此题:给出一个数组,每个元素出现两次除了一个只出现一次,找到出现…

645. Set Mismatch The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another nu…

学习一种知识,我喜欢看看源码是怎么进行它们类之间的关系以及方法的调用,是怎么实现的.这样我才感觉踏实. 既然现在谈到HandlerMapping,我们先知道HandlerMapping的作用:HandlerMapping的作用就是解析请求链接,然后根据请求链接找到执行这个请求的类(HandlerMapping所说的handler,也就是我们写的Controller或是Action). 现在我们来了解HandlerMapping的继承体系图: 至于我们在配置文件中配置的BeanNameUrlHan…

Ok, Go ahead. 1 (a) (b) (c) (d) 2 注:union 在 Common Lisp 中的作用就是求两个集合的并集.但是这有一个前提,即给的两个列表已经满足集合的属性了.具体的操作过程似乎是对第一个 list 中的每一个元素在第二个 list 中查找,如无则标记一下:待第一个 list 的所有元素在第二个 list 中查完以后将所有标记过的元素放入一个 list 中与第二个 list 进行合并.这意味着,如果刚开始给的两个 list 不完全满足集合的属性,则会有重复出现…

Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. 求二叉树的最大深度问题用到深度优先搜索DFS,递归的完美应用,跟求二叉树的最小深度问题原理相同.代码如下: C++ 解法一: class Solution { public: in…