Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1444    Accepted Submission(s): 329 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

Hdu 5274 Dylans loves tree (树链剖分模板) 题目传送门 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <vector> #define ll long…

Dylans loves tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1915    Accepted Submission(s): 492 Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes…

Dylans loves tree Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes on tree is numbered by 1∼N. Then he is given Q questions like that: ①0 x y:change node x′s value to y ②1 x y:For all the value in the path fr…

Dylans loves tree http://acm.hdu.edu.cn/showproblem.php?pid=5274 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i] .Nodes on tree…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5274 [题目大意] 给出一棵树,每个点有一个权值,权值可修改,且大于等于0,询问链上出现次数为奇数的数,题目保证每次询问的链上最多只有一个数出现次数为奇数.如果不存在这样的数,就输出-1. [题解] 题目等价于求链上点的异或和,树链剖分,线段树维护区间异或和,然后链上查询即可,注意到存在权值为0的特殊情况,所以我们将更新的数字都+1,在最后处理答案的时候-1即可. [代码] #include <…

Problem Description Dylans is given a tree with N nodes. All nodes have a value A[i].Nodes on tree is numbered by 1∼N. Then he is given Q questions like that: ①0 x y:change node x′s value to y ②1 x y:For all the value in the path from x to y,do they…

Dylans loves tree view code#pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <vector> #include <queue> #include <…

Pog and Szh are playing games. Firstly Pog draw a tree on the paper. Here we define 1 as the root of the tree.Then Szh choose some nodes from the tree. He wants Pog helps to find the least common ancestor (LCA) of these node.The question is too diffi…

题目地址:HDU 5266 这题用转RMQ求LCA的方法来做的很easy,仅仅须要找到l-r区间内的dfs序最大的和最小的就能够.那么用线段树或者RMQ维护一下区间最值就能够了.然后就是找dfs序最大的点和dfs序最小的点的近期公共祖先了. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm>…