SPOJ - GSS1:https://vjudge.net/problem/SPOJ-GSS1 参考:http://www.cnblogs.com/shanyr/p/5710152.html?utm_source=itdadao&utm_medium=referral 题意: 给定一个数列,很多次询问,问某个区间中最大的连续和是多少. 思路 线段树,每个线段树的节点要维护对应区间的最大值ans,与左端点相连的最大值lv,与右端点相连的最大值rv,还有区间全部的总和V: 这个V用在pushup中…

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…

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 ≤ j ≤ y }. Given M queries, your program must output the results of these…

[题目分析] 线段树裸题. 注意update的操作,写结构体里好方便. 嗯,没了. [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set> #include <queue> #include <string> #include <iostream&…

题目链接:点击打开链接 每一个点都是最大值,把一整个序列和都压缩在一个点里. 1.普通的区间求和就是维护2个值,区间和Sum和延迟标志Lazy 2.Old 是该区间里出现过最大的Sum, Oldlazy 是对于给下一层的子区间的标志,添加多少是能给子区间添加的值最大的(用来维护Old) 显然对于Old .要么维持原样,要么更新为稍新的值:即 Sum(id) + Oldlazy 而对于Oldlazy, 要么维持原样,要么变成最新的延迟标记:即 Lazy(id) + Oldlazy 上2行的Oldl…

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…

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,…

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 output the results of these queries. Inp…

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…

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…