法一:暴力! 让干什么就干什么,那么久需要可持久化线段树了. 但是空间好紧.怎么破? 不down标记好了! 每个点维护sum和add两个信息,sum是这段真实的和,add是这段整体加了多少,如果这段区间被完全包含,返回sum,否则加上add * 询问落在这段区间的长度再递归回答. 怎么还是MLE? 麻辣鸡指针好像8字节,所以改成数组的就过了... #include<cstdio> #include<cstring> #include<cstdlib> #include&

To the moon Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 8372    Accepted Submission(s): 1986 Problem Description BackgroundTo The Moon is a independent game released in November 2011, it is

Distance on the tree DSM(Data Structure Master) once learned about tree when he was preparing for NOIP(National Olympiad in Informatics in Provinces) in Senior High School. So when in Data Structure Class in College, he is always absent-minded about

题目描述 一个长度为n的序列a,设其排过序之后为b,其中位数定义为b[n/2],其中a,b从0开始标号,除法取下整.给你一个长度为n的序列s.回答Q个这样的询问:s的左端点在[a,b]之间,右端点在[c,d]之间的子序列中,最大的中位数.其中a<b<c<d.位置也从0开始标号.我会使用一些方式强制你在线. 输入 第一行序列长度n.接下来n行按顺序给出a中的数. 接下来一行Q.然后Q行每行a,b,c,d,我们令上个询问的答案是x(如果这是第一个询问则x=0). 令数组q={(a+x)%n,

题意: 给出一个具有N个点的树,现在给出两种操作: 1.get x,表示询问以x作为根的子树中,1的个数. 2.pow x,表示将以x作为根的子树全部翻转(0变1,1变0). 思路:dfs序加上一个线段树区间修改查询. AC代码: #include<iostream>#include<vector>#include<string.h>using namespace std;const int maxn=2e5+5;int sum[maxn<<2],lazy[

Problem Description In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.Now Pudge wants to do some operations on th

Mayor's posters POJ 2528 传送门 线段树区间修改加离散化 #include <cstdio> #include <iostream> #include <queue> #include <cmath> #include <algorithm> #include <cstring> using namespace std; #define ll long long #define P pair<int,in

Super Mario Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9618    Accepted Submission(s): 4074 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability

Kth number Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16941    Accepted Submission(s): 5190 Problem Description Give you a sequence and ask you the kth big number of a inteval.   Input The

题意  给出一段初始化全为1的区间  后面可以一段一段更改成 1 或 2 或3 问最后整段区间的和是多少 思路:标准线段树区间和模版题 #include<cstdio> #include<algorithm> #include<set> #include<vector> #include<cstring> #include<iostream> using namespace std; ; struct Node{ int l,r,su

描述 A lot of battleships of evil are arranged in a line before the battle. Our commander decides to use our secret weapon to eliminate the battleships. Each of the battleships can be marked a value of endurance. For every attack of our secret weapon,

Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for

题目 给定一组数,要求进行若干次操作,这些操作可以分为两种类型: (1) CMD 1 beg end value 将数组中下标在[beg, end] 区间内数字都变为value (2) CMD 2 beg end 求出数组中下标在[beg ,end]区间中的所有数字的和 分析 树状数组在区间查询和单点修改情况下效率较线段树高一些,而无法像线段树一样在O(logN)的时间内完成区间修改.因此使用线段树解决.使用线段树主要的是定义好线段树节点的状态.(如代码中注释所说,状态一定要明确,且容易计算!)

和线段树类似,每个结点也要打lazy标记 但是lazy标记和线段树不一样 具体区别在于可持久化后lazy-tag不用往下传递,而是固定在这个区间并不断累加,变成了这个区间固有的性质(有点像分块的标记了) update就按照这么来 int update(int last,int L,int R,int c,int l,int r){ int now=++size; T[now]=T[last]; if(L<=l && R>=r){ T[now].sum+=(r-l+)*c; T[

这就是很简单的基本的线段树的基本操作,区间修改,区间查询,对区间内部信息打上laze标记,然后维护即可. 我自己做的时候太傻逼了...把区间修改写错了,对给定区间进行修改的时候,mid取的是节点的左右的中间值,而不是更新区间的中间值(太菜了). #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ; struct no

https://cn.vjudge.net/problem/HDU-3974 题意 有一棵树,给一个结点分配任务时,其子树的所有结点都能接受到此任务.有两个操作,C x表示查询x结点此时任务编号,T x y表示给x结点分配编号为y的任务. 分析 题目读起来就很有区间修改的味道,将一个区间变为一个值.问题在于怎么把这棵树对应到区间上. 对于一个结点,其控制的范围是它的子树,对应区间范围可以看作是以dfs序表示的区间.好像有点绕..就是给每个结点再对应一个dfs序,然后在dfs时把这个点控制的子树看

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

[SDOI2011]染色 Time Limit: 20 Sec  Memory Limit: 512 MBSubmit: 6870  Solved: 2546[Submit][Status][Discuss] Description 给定一棵有n个节点的无根树和m个操作,操作有2类: 1.将节点a到节点b路径上所有点都染成颜色c: 2.询问节点a到节点b路径上的颜色段数量(连续相同颜色被认为是同一段),如“112221”由3段组成:“11”.“222”和“1”. 请你写一个程序依次完成这m个操作

Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval. In

http://poj.org/problem?id=3468 题目大意: 给你N个数还有Q组操作(1 ≤ N,Q ≤ 100000) 操作分为两种,Q A B 表示输出[A,B]的和   C A B X表示把[A,B]的所有数加上X 思路: 线段树的区间修改..... 昨天晚上改了老半天. 然后关机准备睡觉毕竟今天有实验..去洗个头..突然有灵感..急急忙忙的开电脑改了就对了~哈哈哈 PS:POJ AC 100了~ 因为混迹各个OJ,SO才100 用位运算优化*2  1600+MS,不用2200