[比赛链接] https://codeforces.com/contest/1006 [题解] Problem A. Adjacent Replacements        [算法] 将序列中的所有偶数替换为奇数即可 时间复杂度 : O(N) [代码] #include<bits/stdc++.h> using namespace std; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y);…

D. Arpa's letter-marked tree and Mehrdad's Dokhtar-kosh paths CF741D 题意: 一棵有根树,边上有字母a~v,求每个子树中最长的边,满足这个边上的所有字母重拍后可以构成回文 发明者自己出的题...orz 由于本来知道就是dsu on tree,所以还是想出来了 首先点分治是没法做了,这是有根树 写成二进制,两条链合起来构成回文\(\rightarrow\)异或和为0或者只有一位是1 一开始困惑于只处理到当前根的异或和的话,随着当前…

CF741D Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths 好像这个题只能Dsu On Tree? 有根树点分治 统计子树过x的路径 奇偶可以xor,深度可以减,所以,w[x]x到根的链上二进制数S保留字符出现奇偶性 mx[S]表示w[x]=S的x的最大深度 类比点分治去做 更新答案时候处理一个轻儿子回来更新mx[] 重儿子贡献的答案额外处理. #include<bits/stdc++.h> #define reg reg…

题目链接:Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths 第一次写\(dsu\ on\ tree\),来记录一下 \(dsu\ on\ tree\)主要维护子树信息,往往可以省掉一个数据结构的启发式合并.大体思路如下: 轻重链路径剖分之后,对每个点先递归处理他的所有轻儿子,每次处理完轻儿子之后把这棵子树的信息清空.最后再来处理重孩子,重儿子的信息就可以不用清空了.由于我们是用一个全局数组来记录信息的,重儿子子树的信息就仍然保留…

地址:http://codeforces.com/contest/766/problem/E 题目: E. Mahmoud and a xor trip time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Mahmoud and Ehab live in a country with n cities numbered from …

本文版权归ljh2000和博客园共有,欢迎转载,但须保留此声明,并给出原文链接,谢谢合作. 本文作者:ljh2000 作者博客:http://www.cnblogs.com/ljh2000-jump/转载请注明出处,侵权必究,保留最终解释权! Description Mahmoud and Ehab live in a country with n cities numbered from 1 to n and connected by n - 1 undirected roads. It's…

Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime? Example: Input: [3, 10, 5, 25, 2, 8] Output: 28 Explanation: The maximum resul…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

题目链接 XOR 游戏 Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 256    Accepted Submission(s): 86 Problem Description众所周知,度度熊喜欢XOR运算[(XOR百科)](http://baike.baidu.com/view/674171.htm). 今天,它发明了一种XOR新游戏…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 思路:用两个stack<TreeNode*> in , s; in : 记录当前的路径 p  , 和vector<…

UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径. f[i][j][k]从下往上到第i层第j个和为k的方案数 上下转移不一样,分开处理 没必要判断走出沙漏 打印方案倒着找下去行了,尽量往左走   沙茶的忘注释掉文件WA好多次   #include <iostream> #include <cstdio> #include <a…

Description Input 第一行包含两个整数N和 M, 表示该无向图中点的数目与边的数目. 接下来M 行描述 M 条边,每行三个整数Si,Ti ,Di,表示 Si 与Ti之间存在 一条权值为 Di的无向边. 图中可能有重边或自环. Output 仅包含一个整数,表示最大的XOR和(十进制结果),注意输出后加换行回车. 这道题好像是很久以前学线性基的时候留下的--现在来填个坑-- 首先,由于异或有一个很好的性质,就是两个相同的数异或起来等于零.所以,一条边重复走两遍不会对答案产生贡献.这…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

前段时间的一次样本,需要给出专杀,应急中遇到的是linux中比较常见的两个家族gates和xor. 首先是xor的专杀脚本,xor样本查杀的时候需要注意的是样本的主进程和子进程相互保护(详见之前的xor ddos分析http://www.cnblogs.com/goabout2/p/4888651.html),想要杀掉的话,需要先通过kill –stop挂起主进程,再删除其他的文件,但是由于xor的进程名是随机值,同时主机上还有有gates木马(gates最显著的特征就是会替换系统文件ps,ls…

Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middl…

Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13717   Accepted: 5824 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the…

题目描述: 1015. Jill's Tour Paths Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description Every year, Jill takes a bicycle tour between two villages. There are different routes she can take between these villages, but she does have an upper limit…

XOR and Favorite Number time limit per test: 4 seconds memory limit per test: 256 megabytes input: standard input output: standard output Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pa…

这题在Unique Paths的基础上增加了一些obstacle的位置,应该说增加的难度不大,但是写的时候对细节的要求多了很多,比如,第一列的初始化会受到之前行的第一列的结果的制约.另外对第一行的初始化,也要分if else赋值.很容易出现初始化不正确的情况. 代码: class Solution { public: int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) { ][]==) ; int…

#include <cstdio> #include <cstring> ; ; int cnt,Ans,b,x,n; inline int Max(int x,int y) {return x>y?x:y;} ];}Tree[Maxn*Len]; void Insert(int x) { ; bool k; ;i--) { k=x&(<<i); ) Tree[Now].next[k]=++cnt; Now=Tree[Now].next[k]; } } i…

唯一路径问题II Unique Paths II Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is…

http://acm.hdu.edu.cn/showproblem.php?pid=5661 Claris and XOR Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 146    Accepted Submission(s): 51 Problem Description Claris loves bitwise operatio…

https://leetcode.com/problems/unique-paths/ 这道题,不利用动态规划基本上规模变大会运行超时,下面自己写得这段代码,直接暴力破解,只能应付小规模的情形,当23*12时就超时了: class Solution { public: // Solution():dp1(m,vector<int>(n,-1)),dp2(m,vector<int>(n,-1)){ // } int uniquePaths(int m, int n) { helper…

异或链表(Xor Linked List)也是一种链式存储结构,它可以降低空间复杂度达到和双向链表一样目的,任何一个节点可以方便的访问它的前驱节点和后继结点.可以参阅wiki 普通的双向链表 class Node { public: int data; Node *prev; Node *next; }; class BiLinkedList { public: Node *head; Node *tail; }; 普通双向链表的一个节点表示如下: 完整的普通双向链表如下所示:   对于异或链表…

Claris and XOR Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 744    Accepted Submission(s): 330 Problem Description Claris loves bitwise operations very much, especially XOR, because it has ma…

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in t…

题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=2819 分析: 树上的nim游戏,关键就是要判断树上的一条链的异或值是否为0 这个题目有单点修改和一条链上的询问,很显然可以用树链剖分做,但是n<=50W,所以会TLE+爆栈 我们设f[x]表示点x到根节点这条路径上的异或和,那么很显然如果询问u,v上的异或和,那么ans=f[u]^f[v]^v[lca(u,v)] 那么接下来的问题就是对于每个单点修改,如何维护f[] 容易看出来,如果…

计算1到n的一条路径使得路径上的值xor和最大. 先任意走一条路径计算xor和,然后dfs的时候处理出所有的环的xor和,这样对于所有的环的xor和求线性基,在任意走出的路径的xor和上贪心即可. 正确性显然,如果环与选择的路径有重合,那么重合的部分就会被xor两次,也就没有xor,相当于更改了一部分路径.如果环与选择的路径没有重合,那么相当于从路径上任意一个点到环上的一个点,跑一圈后从进入环的点原路返回,这样环的xor和就计算到了,而往返两次的路径也因为xor了两次相当于没有xor,就不用考虑…