虽然每次给一个区间,但是可以看作在区间内进行数个点操作,同样数列下标是动态变化的,如果我们将每个字符出现看作1,被删除看作0,则通过统计前缀和就能轻松计算出两个端点的位置了!这正是经典的树状数组操作 需要注意的是,即使每次找到了两个端点,如果只是朴素总左到右遍历会TLE,所以可以给每个字符开一个set,每个set储存字符出现的位置. #include <iostream> #include <cstdio> #include <cstring> #include <…

F. Letters Removing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Petya has a string of length n consisting of small and large English letters and digits. He performs m operations. Each…

Codeforces 899F - Letters Removing 思路:考虑一下怎么找到输入的l和r在原来串中的位置,我们想到用前缀和来找,一开始所有位置都为1,删掉后为0,那么前缀和为l的位置就是l的位置,前缀和为r的位置就是r的位置. 那么用树状数组来维护一下前缀和,再用二分查找来找到位置就可以了.然后找到的区间可能很大,然而我们要删的只有一种字符,把每种字符的位置分开来保存来操作,可以用set也可以用vector. 代码1(set版,187ms): #include<bits/stdc…

Letters Removing 题意:给你一个长度为n的字符串,然后进行m次删除操作,每次删除区间[l,r]内的某个字符,删除后并且将字符串往前补位,求删除完之后的字符串. 题解:先开80个set 将每个字符对应的下标存入空间, 然后每次删除了一个字符之后就将字符串的相应位置改成一个不会产生干扰的字符(我这里使用的是'.'). 并且在线段树的相应位置标记一下.然后每次删除时候的左右区间就用2分区查找. 找到位置pos 使得 [1,pos]的值等于 pos - l, 这样就可以找到相应的区间了.…

The best programmers of Embezzland compete to develop a part of the project called "e-Government" — the system of automated statistic collecting and press analysis. We know that any of the k citizens can become a member of the Embezzland governm…

题目链接:Letters Removing 题意: 给你一个长度为n的字符串,给出m次操作.每次操作给出一个l,r和一个字符c,要求删除字符串l到r之间所有的c. 题解: 看样例可以看出,这题最大的难点在于每次在字符串中删除了前面的字符会对后面的字符产生影响.如何确定当前l和r所指的字符?这里由于对字符的位置查询相当于单点操作区间查询,可以用树状数组维护字符串的前缀和,这样就可以确定l和r的位置了(二分+树状数组 : 复杂度(log(n)×log(n))).再把所有的字符放到set里面进行删除操…

Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 100 000, inclusive. It is possible that some cards have the same integers on them. Vasily decided to sort the cards. To do this,…

C. Propagating tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/383/problem/C Description Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numb…

传送门:https://codeforces.com/contest/540/problem/E 题意: 有一段无限长的序列,有n次交换,每次将u位置的元素和v位置的元素交换,问n次交换后这个序列的逆序对个数为多少 题解: 因为值域范围为1e9,而n的范围只有1e5,所以我们肯定是不能直接交换的,对n次操作离散化,离散化后的数组最大为2e5,这里需要用到一些离散化的技巧. 将每次交换的u,v两个点放到map里面,键为pos,值为0 然后对于map映射,值就是离散化后的下标 离散化后我们应该做什么…

Camping Groups 题目连接: http://codeforces.com/problemset/problem/173/E Description A club wants to take its members camping. In order to organize the event better the club directors decided to partition the members into several groups. Club member i has…

题目链接: D. Pashmak and Parmida's problem time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her…

E. Hanoi Factory time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Of course you have heard the famous task about Hanoi Towers, but did you know that there is a special factory producing the…

B. Cards Sorting time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this integer is between 1 and 10…

题目:戳这里 学习博客:戳这里 题意:有很多个活动,每个活动有持续天数,每个活动会在每天提供C个CPU每个CPU价格为P,问需要工作N天,每天需要K个CPU的最少花费. 解题思路:遍历每一天,维护当前天K个cpu的最小花费.具体方法是维护两个线段树(树状数组也可以),维护每一天可以使用的cpu数和价格*cpu数的前缀和.注意数组下标是价格(1e6的数组. (不明白的话可以看代码,代码思路很清晰 附学习博客的代码: 1 #include <iostream> 2 3 #include <a…

题目链接:http://codeforces.com/contest/828/problem/E E. DNA Evolution time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Everyone knows that DNA strands consist of nucleotides. There are four typ…

题目链接:http://codeforces.com/problemset/problem/652/D 大意:给若干个线段,保证线段端点不重合,问每个线段内部包含了多少个线段. 方法是对所有线段的端点值离散化,按照左端点从大到小排序,顺着这个顺序处理所有线段,那么满足在它内部的线段一定是之前已经扫到过的.用树状数组判断有多少是在右端点范围内. #include <iostream> #include <vector> #include <algorithm> #incl…

http://codeforces.com/problemset/problem/992/E 题意:给定一个序列 ai​ ,记其前缀和序列为 si​ ,有 q 个询问,每次单点修改,询问是否存在一个 i 满足 ai​=si−1​,有多解输出任意一个,无解输出 -1. 思路一:构造一个b[i] = a[i] - s[i - 1]的序列,答案就是在其中寻找为0的位置,对每一次操作进行一个线段树的单点修改和区间修改之后对一整个区间寻找是否存在0的位置,但是最坏的情况下能达到N * Q * lnN,虽然…

E. DNA Evolution 题目连接: http://codeforces.com/contest/828/problem/E Description Everyone knows that DNA strands consist of nucleotides. There are four types of nucleotides: "A", "T", "G", "C". A DNA strand is a seque…

Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other. The boys decided to h…

D. Nested Segments 题目连接: http://www.codeforces.com/contest/652/problem/D Description You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains. Input The first line…

http://codeforces.com/problemset/problem/570/D Tree Requests time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Roman planted a tree consisting of n vertices. Each vertex contains a lowercase…

任意门:http://codeforces.com/contest/652/problem/D D. Nested Segments time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are given n segments on a line. There are no ends of some segments th…

链接:http://codeforces.com/problemset/problem/570/D D. Tree Requests time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Roman planted a tree consisting of n vertices. Each vertex contains a low…

题目链接:http://codeforces.com/problemset/problem/216/D 题意: 对于一个梯形区域,假设梯形左边的点数!=梯形右边的点数,那么这个梯形为红色.否则为绿色, 问: 给定的蜘蛛网中有多少个红色. 2个树状数组维护2个线段.然后暴力模拟一下,由于点数非常多但须要用到的线段树仅仅有3条,所以类似滚动数组的思想优化内存. #include<stdio.h> #include<iostream> #include<string.h> #…

CodeForces - 369E Valera and Queries 题目大意:给出n个线段(线段的左端点和右端点坐标)和m个查询,每个查询有cnt个点,要求给出有多少条线段包含至少其中一个点. 思路:如果按照题意正面去算有每个线段是否包含点,那么时间复杂度是不允许的:如果去算每个点被哪些线段包含,那么去重比较困难.正难则反,因此对于每个查询,选择去求有多少个线段没有覆盖任何一个点,那么答案则为n减所求.用树状数组来统计没有覆盖任何一个点的线段数,大致做法是将临界线段(如x1+1,x2-1)…

链接:https://codeforces.com/contest/1269/problem/E 题意:给一个序列P1,P2,P3,P4....Pi,每次可以交换两个相邻的元素,执行最小次数的交换移动,使得最后存在一个子段1,2,…,k,这是题目所定义的f(k),题目要求求出所有的f(n),并依次输出. 思路:首先考虑逆序对问题,比如3 2 1 4这个序列,要使其变为1 2 3 4,最小的移动次数是这个序列中逆序对之和,2+1 = 3,逆序对是(3,2) (3,1)(2,1),但是在比如序列3…

Dynamic Rankings Time Limit: 10 Seconds      Memory Limit: 32768 KB The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They…

HDU 3333:http://acm.hdu.edu.cn/showproblem.php?pid=3333 这两个题是类似的,都是离线处理查询,对每次查询的区间的右端点进行排序.这里我们需要离散化处理一下,标记一下前面是否出现过这个值,然后不断更新last数组(该数组保存的是每个数最后一次出现的位置).最后用树状数组维护. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ; struct node{…

1.HDU 1556  Color the ball   区间更新,单点查询 2.题意:n个气球,每次给(a,b)区间的气球涂一次色,问最后每个气球各涂了几次. (1)树状数组 总结:树状数组是一个查询和修改复杂度都为log(n)的数据结构.主要用于查询任意两位之间的所有元素之和,但是每次只能修改一个元素的值. 这里改下思路可以用树状数组.在更新(a,b)时,向上更新,将a~n加1,b+1~n减1.查询点时,向下求和即可. #include<iostream> #include<cstr…

HDU 3966 Aragorn's Story 先把树剖成链,然后用树状数组维护: 讲真,研究了好久,还是没明白 树状数组这样实现"区间更新+单点查询"的原理... 神奇... #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #inc…