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 placing the…

poj2528 Mayor's posters 题意:在墙上贴海报,海报可以互相覆盖,问最后可以看见几张海报 思路:这题数据范围很大,直接搞超时+超内存,需要离散化: 离散化简单的来说就是只取我们需要的值来用,比如说区间[1000,2000],[1990,2012] 我们用不到[-∞,999][1001,1989][1991,1999][2001,2011][2013,+∞]这些值,所以我只需要1000,1990,2000,2012就够了,将其分别映射到0,1,2,3,在于复杂度就大大的降下来了…

Mayor's posters Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 37346Accepted: 10864 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at…

Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 43507   Accepted: 12693 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral post…

Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 39795   Accepted: 11552 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral post…

Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 50888   Accepted: 14737 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral post…

题目链接:https://vjudge.net/problem/POJ-2528 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…

Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 70365   Accepted: 20306 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral post…

线段树 + 离散化 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…

题目描述 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 placing…

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 placing the…

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…

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…

题目传送门: POJ-2528 题意就是在一个高度固定的墙面上贴高度相同宽度不同的海报,问贴到最后还能看到几张?本质上是线段树区间更新问题,但是要注意的是题中所给数据范围庞大,直接搞肯定会搞出问题,所以要离散化,而离散化的过程中要注意一个问题,比方说1-10,1-5,6-10,本来是可以三张海报都可以看见的,但是按照题意来看是看不到的,因为他是一个点代表一个单位长度(诡异>_<),解决的办法则是对于距离大于1的两相邻点,中间再插入一个点...... 代码如下,我用了vector,和用数组的相差…

给一块最大为10^8单位宽的墙面,贴poster,每个poster都会给出数据 a,b,表示该poster将从第a单位占据到b单位,新贴的poster会覆盖旧的,最多有10^4张poster,求最后贴完,会看到几张poster (哪怕只露出一个单位,也算该poster可见): 我一看这么大数据,又看了下时间限制只有1s,不科学啊,如果真的按10^8建树不可能过时间啊,而且根据它的空间限制,大概只能建10^7这么大的数组. 后来搜博客发现大家的标题都写着离散化,原来用离散化做这个题目,但是我不会离…

http://poj.org/problem?id=2528 https://www.luogu.org/problem/UVA10587 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…

题目分析:线段树区间更新+离散化 代码如下: # include<iostream> # include<cstdio> # include<queue> # include<vector> # include<list> # include<map> # include<set> # include<cstdlib> # include<string> # include<cstring&g…

恩,这区间范围挺大的,需要离散化.如果TLE,还需要优化一下常数. AC代码 #include <stdio.h> #include <string.h> #include <map> #include <set> #include <algorithm> using namespace std; +; typedef pair<int, int> Pii; Pii a[ + ]; +], c[maxn]; ]; void build…

题意:给一个区间,表示这个区间贴了一张海报,后贴的会覆盖前面的,问最后能看到几张海报. 思路: 之前就不会离散化,先讲一下离散化:这里离散化的原理是:先把每个端点值都放到一个数组中并除重+排序,我们就得到了处理后的数组,现在我们只需要用二分查找端点值在整个数组的下标,这样就达到了离散化的目的,压缩了长度.因为这里很特殊,不能用一般的离散化去做,如果区间两端只差1,那么需要给这个区间再加一个值,这个其他题解讲到了. 这里的区间更新和上一题不太一样,有一些地方要注意一下 代码: #include<q…

主要是参考了这个博客 地址戳这儿 题目大意:n(n<=10000) 个人依次贴海报,给出每张海报所贴的范围li,ri(1<=li<=ri<=10000000) .求出最后还能看见多少张海报. 对于输入要离散化,之后线段树维护张贴的状态. 留意由于一般的离散化可能导致一些区间被“挤压掉” 比如 1-10 1-4 5-101-10 1-4 6-10  被处理成了一样的情况导致错漏. 方法是在相差超过1的相邻点插多一个点,使得离散化的时候不会挤压在一起.还有要注意的地方就是空间开多大,一…

题目大意:有t组数据,每组数据给你n张海报(1<=n<=10000),下面n组数据分别给出每张海报的左右范围(1 <= l <= r <= 10000000),下一张海报会覆盖前一张海报,求最后可见(包括完全和不完全可见)的海报有几张. 例如: 1 5 1 4 2 6 8 10 3 4 7 10 如上图所示,答案为4. 解题思路:其实这是一道区间染色问题,但是由于查找区间太大,显然直接建树会导致MLE,所以这里通过使用对区间的离散化来缩小查找范围.参考了一些大牛博客,简单说一…

题意 : 在墙上贴海报, n(n<=10000)个人依次贴海报,给出每张海报所贴的范围li,ri(1<=li<=ri<=10000000).求出最后还能看见多少张海报. 分析 : 很容易想到利用线段树来成段置换,最后统计总区间不同数的个数.但是这里有一个问题,就是区间可以很大,线段树开不了那么大的空间,遂想能不能离散化.实际上只记录坐标的相对大小进行离散化最后是不影响我们计算的,但是光是普通的离散化是不行的,就是我们贴海报的实际意义是对(l, r)段进行添加,而不是对于这个区间的点…

离散化的思想: 对于这样的数据 (3,10000). (9,1000000). (5.100000), (1,1000). (7,1000000) 我们能够将其处理为 (2,7). (5,9). (3,8), (1,6), (4,9) 我们再对离散化之后的数据进行处理即可了. 题目意思: n(n<=10000)个人依次贴海报,给出每张海报所贴的范围li,ri(1<=li<=ri<=10000000). 求出最后还能看见多少张海报. 參考代码: #include <iostre…

[poj2528]Mayor's posters Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 66154   Accepted: 19104 Description The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their elect…

POJ2528.Mayor's posters 这道题真的是线段数的经典的题目,因为数据很大,直接建树的话肯定不可以,所以需要将数据处理一下,没有接触离散化的时候感觉离散化这个东西相当高级,其实在不知道离散化是什么东西之前,就已经用过这种东西了,只是不知道叫什么.关于离散化,就是根据数的相对大小对他们进行映射,用小的数来表示他们.所以要先排序,然后去重,然后操作就可以.有排序版的离散化和STL版的离散化(自己起的名字 ...),代码里写的是排序版的,就是for循环遍历的时候,和身边的数比较一下是…

转载自:http://blog.csdn.net/shiqi_614/article/details/8228102 之前做了些线段树相关的题目,开学一段时间后,想着把它整理下,完成了大牛NotOnlySuccess的博文“完全版线段树”里的大部分题目,其博文地址Here,然后也加入了自己做过的一些题目.整理时,更新了之前的代码风格,不过旧的代码仍然保留着. 同样分成四类,不好归到前四类的都分到了其他.树状数组能做,线段树都能做(如果是内存限制例外),所以也有些树状数组的题目,会标示出来,并且放…

原文出处:http://www.notonlysuccess.com/ (好像现在这个博客已经挂掉了,在网上找到的全部都是转载) 今天在清北学堂听课,听到了一些很令人吃惊的消息.至于这消息具体是啥,等我晚上再说23333 我大概的看了一下所讲内容及实例,果不愧是大牛级别的人,讲的很详细,所以决定转载过来. 对原文作者致以崇高的敬意. 以下是转载内容. [完全版]线段树 很早前写的那篇线段树专辑至今一直是本博客阅读点击量最大的一片文章,当时觉得挺自豪的,还去pku打广告,但是现在我自己都不太好意思…

[POJ 3468]A Simple Problem with Integers给定Q个数A1, ..., AQ,多次进行以下操作:1.对区间[L, R]中的每个数都加n.2.求某个区间[L, R]中的和.Q ≤ 100000 Sol如果只记录区间的和?进行操作1的时候需要O(N)的时间去访问所有的节点.考虑多记录一个值inc,表示这个区间被整体的加了多少. 延迟更新信息更新时,未必要真的做彻底的更新,可以只是将应该如何更新记录下来,等到真正需要查询准确信息时,才去更新 足以应付查询的部分.在区…

线段树完全版  ~by NotOnlySuccess 很早前写的那篇线段树专辑至今一直是本博客阅读点击量最大的一片文章,当时觉得挺自豪的,还去pku打广告,但是现在我自己都不太好意思去看那篇文章了,觉得当时的代码风格实在是太丑了,很多线段树的初学者可能就是看着这篇文章来练习的,如果不小心被我培养出了这么糟糕的风格,实在是过意不去,正好过几天又要给集训队讲解线段树,所以决定把这些题目重新写一遍,顺便把近年我接触到的一些新题更新上去~;并且学习了splay等更高级的数据结构后对线段树的体会有更深了一…

简单数据结构 本节课可能用到的一些复杂度: O(log n). 1/1+1/1/.....1/N+O(n log n) 在我们初学OI的时候,总会遇到这么一道题. 给出N次操作,每次加入一个数,或者询问当前所有数的最大值. 维护一个最大值Max,每次加入和最大值进行比较.(这其实就是一个冒泡排序) 简单的代码实现一下 ;i<=n;++i) { MAX=max(MAX,a[i]); } 时间复杂度是O(N) EX:入门题 给出N次操作,每次加入一个数,删除一个之前加入过的数Ai,或者询问当前所有数…