GSS1 线段树最大子段和裸题,不带修改,注意pushup. 然而并不会猫树之类的东西 #include<bits/stdc++.h> #define MAXN 50001 using namespace std; struct node{ int l , r , sum , lMax , rMax , midMax; }Tree[MAXN << ]; int a[MAXN] , rMax , allMax , N; inline int max(int a , int b){ r…
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…
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…
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…
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…
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…