GSS7Can you answer these queries VII 给出一棵树,树的节点有权值,有两种操作: 1.询问节点x,y的路径上最大子段和,可以为空 2.把节点x,y的路径上所有节点的权值置为c 分析: 修改树路径的信息,可以考虑一下树链剖分.动态树. 这题可以用树链剖分的方式来做,不会的可以看看这篇 树链剖分---模板.其实树链剖分不难理解,一小时左右就能学会了. 对于在一段区间的最大子段和问题,可以参考GSS1 spoj 1043 Can you answer these qu…

题目链接:http://www.spoj.com/problems/QTREE/en/ QTREE - Query on a tree #tree You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form:…

[题目分析] 垃圾vjudge又挂了. 树链剖分裸题. 垃圾spoj,交了好几次,基本没改动却过了. [代码](自带常数,是别人的2倍左右) #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 20005 int T,n,fr[maxn],h[maxn],to[maxn],ne[maxn]…

[题目分析] 问题放到了树上,直接链剖+线段树搞一搞. 调了300行+. (还是码力不够) [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set> #include <queue> #include <string> #include <iostr…

Query on a tree again! 给出一棵树,树节点的颜色初始时为白色,有两种操作: 0.把节点x的颜色置反(黑变白,白变黑). 1.询问节点1到节点x的路径上第一个黑色节点的编号. 分析: 先树链剖分,线段树节点维护深度最浅的节点编号. 注意到,如果以节点1为树根时,显然每条重链在一个区间,并且区间的左端会出现在深度浅的地方.所以每次查找时发现左区间有的话,直接更新答案. 9929151 2013-08-28 10:45:55 Query on a tree again! 100…

题意:一棵包含N 个结点的树,每条边都有一个权值,要求模拟两种操作:(1)改变某条边的权值,(2)询问U,V 之间的路径中权值最大的边. 思路:最近比赛总是看到有树链剖分的题目,就看了论文,做了这题,思路论文上讲的很清楚了,好长时间没写线段树了,错了好几遍.对树进行轻重边路径剖分.对于询问操作,我们可以分别处理两个点到其最近公共祖先的路径.路径可以分解成最多O(log N)条轻边和O(log N)条重路径,那么只需考虑如何维护这两种对象.对于轻边,我们直接处理即可.而对于重路径,我们只需用线段树…

传送门:Problem QTREE https://www.cnblogs.com/violet-acmer/p/9711441.html 题解: 树链剖分的模板题,看代码比看文字解析理解来的快~~~~~~~ AC代码献上: #include<iostream> #include<cstdio> #include<cmath> #include<cstring> using namespace std; #define ls(x) ((x)<<1…

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3-N-1. We will ask you to perfrom some instructions of the following form: CHANGE i ti : change the cost of the i-th edge to ti or QUERY a b : ask for…

Given a tree with N ( N<=100000 ) nodes. Each node has a interger value x_i ( |x_i|<=10000 ). You have to apply Q ( Q<=100000 ) operations: 1. 1 a b : answer the maximum contiguous sum (maybe empty,will always larger than or equal to 0 ) from the…

传送门 题意:给出一个长度为$N$的数列,$Q$次询问,每一次询问$[l,r]$之间的最大子段和,相同的数只计算一次.所有数字的绝对值$\leq 10^5$ GSS系列中不板子的大火题,单独拿出来写 因为相同的数字只计算一次,像GSS1中的合并操作就无法进行,传统做法失效,我们需要一种更强大的做法. 考虑到去重,与HH的项链很相似,所以考虑离线.对询问以$r$从小到大进行排序后进行计算. 考虑到每一次$r$的增加都会产生新的可能的最大子段和,我们用如下方式维护线段树:对于第$i$个叶子节点,它包…

Count on a tree SPOJ 10628 主席树+LCA(树链剖分实现)(两种存图方式) 题外话,这是我第40篇随笔,纪念一下.<(￣︶￣)↗[GO!] 题意 是说有棵树,每个节点上都有一个值,然后让你求从一个节点到另一个节点的最短路上第k小的值是多少. 解题思路 看到这个题一想以为是树链剖分+主席树,后来写着写着发现不对,因为树链剖分我们分成了一小段一小段,这些小段不能合并起来求第k小,所以这个想法不对.奈何不会做,查了查题解,需要用LCA(最近公共祖先),然后根据主席树具有区间加…

SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for giv…

[题解] SPOJ GSS1 - Can you answer these queries I · 题目大意 要求维护一段长度为 \(n\) 的静态序列的区间最大子段和. 有 \(m\) 次询问,每次询问输出区间 \([L,R]\) 的最大子段和. \(|a[i]| \leq 15007\),\(1 \leq m,n\leq5\times10^4\) · 解题思路 首先想到如果用线段树的方法,那么预处理时间复杂度为\(O(n)\),总询问复杂度为\(O(m\cdot logn)\). 当然这么想…

<树链剖分及其应用> 一文讲得非常清楚,我一早上就把他学会了并且A了这题的入门题. spoj QTREE 题目: 给出一棵树,有两种操作: 1.修改一条边的边权. 2.询问节点a到b的最大边权. 直接粘代码.更成熟的代码可以看下一篇BZOJ 1036: [ZJOI2008]树的统计Count #include <set> #include <map> #include <list> #include <cmath> #include <qu…

SPOJ太慢了,SPOJ太慢了, 题意:给定n(n<=10000)个节点的树,每条边有边权,有两种操作:1.修改某条变的边权:2.查询u,v之间路径上的最大边权. 分析:树链剖分入门题,看这里:http://blog.sina.com.cn/s/blog_6974c8b20100zc61.html 轻重链剖分,线段树维护,复杂度 O(nlogn + q* logn * logn ) SPOJ太慢了,我知道我写的应该是没错了, 最后把自己宏定义的max和min去掉之后终于过了,傻逼笑呵呵了..为啥…

You are given a tree with N nodes. The tree nodes are numbered from 1 to N and have colors C1, C2,. . . , CN initially. You have to handle M instructions on the tree of the following forms:• 0 u c: Change the color of node u to c.• 1 u v: Output the…

Can you answer these queries II Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 https://www.spoj.com/problems/GSS2/ Description Being a completist and a simplist, kid Yang Zhe cannot solve but get Wrong Answer from most of the OI problems. And he refuse…

2482: [Spoj1557] Can you answer these queries II Time Limit: 20 Sec  Memory Limit: 128 MBSubmit: 145  Solved: 76[Submit][Status][Discuss] Description 给定n个元素的序列. 给出m个询问:求l[i]~r[i]的最大子段和(可选空子段). 这个最大子段和有点特殊:一个数字在一段中出现了两次只算一次. 比如:1,2,3,2,2,2出现了3次,但只算一次,…

4487. Can you answer these queries VI Problem code: GSS6 Given a sequence A of N (N <= 100000) integers, you have to apply Q (Q <= 100000) operations: Insert, delete, replace an element, find the maximum contiguous(non empty) sum in a given interval…

1557. Can you answer these queries II Problem code: GSS2 Being a completist and a simplist, kid Yang Zhe cannot solve but get Wrong Answer from most of the OI problems. And he refuse to write two program of same kind at all. So he always failes in co…

GSS2 - Can you answer these queries II #tree Being a completist and a simplist, kid Yang Zhe cannot solve but get Wrong Answer from most of the OI problems. And he refuse to write two program of same kind at all. So he always failes in contests. When…

Can you answer these queries VI Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on SPOJ. Original ID: GSS664-bit integer IO format: %lld      Java class name: Main Given a sequence A of N (N <= 100000) integers, you have to appl…

Can you answer these queries I SPOJ - GSS1 You are given a sequence A[1], A[2], -, A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows: Query(x,y) = Max { a[i]+a[i+1]+-+a[j] ; x ≤ i ≤ j ≤ y }. Given M queries, your program must o…

Time Limit: 1000MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Description Being a completist and a simplist, kid Yang Zhe cannot solve but get Wrong Answer from most of the OI problems. And he refuse to write two program of same kind at…

gss5 Can you answer these queries V 给出数列a1...an,询问时给出: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 j <= y2 and x1 <= x2 , y1 <= y2 } 分析: 其实画个图分类讨论一下之后,跟gss1基本一样... 注意到x1<=x2 , y1<=y2. 所以大致可以分为: 1.y1<x2: 直接计…

gss2调了一下午,至今还在wa... 我的做法是:对于询问按右区间排序,利用splay记录最右的位置.对于重复出现的,在splay中删掉之前出现的位置所在的节点,然后在splay中插入新的节点.对于没有出现过的,直接插入.询问时直接统计区间的最大子段和. gss2没能调出bug,所以看了一下以下的gss3,发现跟gss1基本一样.直接上代码 以上的做法是错的,对于这种数据就过不了.姿势不对,囧 44 -2 3 -211 4 GSS Can you answer these queries II…

今天下午不知道要做什么,那就把gss系列的线段树刷一下吧. Can you answer these queries I 题目:给出一个数列,询问区间[l,r]的最大子段和 分析: 线段树简单区间操作. 线段树中记录lx,rx,mx,分别表示:最大前驱连续和,最大后继连续和,区间最大子段和. 在合并时时只需要合并两个区间即可,具体可以看代码的Union. 从队友jingo那里学到了这种合并的写法,发现比网上大部分代码简单很多. #include <set> #include <map&g…

GSS3 - Can you answer these queries III You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations: modify the i-th element in the sequence or for given x y print max{…

Time Limit: 132MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Description You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+…

Time Limit: 115MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu Description You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows: Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i…