Billboard Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2795 Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2795 题目大意:有一块h*w的矩形广告板,要往上面贴广告;   然后给n个1*wi的广告,要求把广告贴上去: 而且要求广告要尽量往上贴并且尽量靠左;  求每个广告的所在的位置,不能贴则为-1. 用线段树模拟,要是左子树的最大值比当前广告大,就查询更新左子树,否则就右子树. #include <iostream> #include <cstdio> #include <cstrin…

手动博客搬家:本文发表于20170822 21:30:17, 原地址https://blog.csdn.net/suncongbo/article/details/77488127 URL: http://acm.hdu.edu.cn/showproblem.php?pid=2795题目大意:有一个h*w的木板 (h, w<=1e9), 现在有n (n<=2e5)张1*xi的海报要贴在木板上,按1~n的顺序每次贴海报时会选择最上的一排的最左边贴 (海报不能互相覆盖), 求每张海报会被贴在哪一行…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11861    Accepted Submission(s): 5223 Problem Description At the entrance to the university, there is a huge rectangular billboard of s…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 21203    Accepted Submission(s): 8750 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions, changes in th…

题意 : 有一块 h * w 的公告板,现在往上面贴 n 张长恒为 1 宽为 wi 的公告,每次贴的地方都是尽量靠左靠上,问你每一张公告将被贴在1~h的哪一行?按照输入顺序给出. 分析 : 这道题说明了每一次贴都尽量选择靠上靠左的位置,那既然这样,我们以1~h建立线段树,给每一个叶子节点赋值初值 w 表示当前行最大能够容纳宽度为 w 的公告纸,那么对于某一输入 wi 只要在线段树的尽量靠左的区间找出能够容纳这张公告的位置(即叶子节点)然后减去 wi 即可,需要对query()函数进行一点改造,将…

题目大意:在h*w 高乘宽这样大小的 board上要贴广告,每个广告的高均为1,wi值就是数据另给,每组数组给了一个board和多个广告,要你求出,每个广告应该贴在board的哪一行,如果实在贴不上,就输出-1: 这个题目也难以想到居然是用线段树来做 我们需要考虑的是,线段树究竟表示的是什么数据.在这个题目里,由于每个广告的高都为1,是不是好像感觉每一行都是一个叶子节点一样.没错,就是这样...化抽象为具体一点,那就是把这个board给竖起来,这样最底部的孩子存贮了当前行的空闲宽度,每个父节点都…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is…

HUD.2795 Billboard ( 线段树 区间最值 单点更新 单点查询 建树技巧) 题意分析 题目大意:一个h*w的公告牌,要在其上贴公告. 输入的是1*wi的w值,这些是公告的尺寸. 贴公告要满足的条件: 1. 尽量往上,同一高度尽量靠左. 2. 求第n个广告所在的行数. 3. 没有合适的位置贴了则输出-1. 建树技巧 首先看了一下数据范围h和w都在1e9,按照高度建树不太现实的.但是发现询问只有2e5,.仔细想一下如果h < 2e5, 我们用h建树,否则用n建树.原因是,每个高度只放…

题目链接 Problem Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is its height and w is its width). The board is the place where all possible announcements are posted: nearest programming competitions,…

HDU 2795 Billboard(宣传栏贴公告,线段树应用) ACM 题目地址:HDU 2795 Billboard 题意:  要在h*w宣传栏上贴公告,每条公告的高度都是为1的,并且每条公告都要尽量贴最上面最靠左边的,给你一系列的公告的长度,问它们能不能贴上. 分析:  不是非常好想,只是想到了就非常好写了.  仅仅要把宣传栏倒过来就好办了,这时候就是变成有h条位置能够填公告,填放公告时就能够尽量找最左边的合适的位置来放了.  能够用线段树实现,查找的复杂度是O(logn),须要注意的坑点…

Attack Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 2496    Accepted Submission(s): 788 Problem Description Today is the 10th Annual of “September 11 attacks”, the Al Qaeda is about to attack…

Coder Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4838    Accepted Submission(s): 1853 Problem Description In mathematics and computer science, an algorithm describes a set of procedures…

Man Down Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2030    Accepted Submission(s): 743 Problem Description The Game “Man Down 100 floors” is an famous and interesting game.You can enjoy t…

任意门:http://acm.hdu.edu.cn/showproblem.php?pid=2795 Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 28743    Accepted Submission(s): 11651 Problem Description At the entrance to the un…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2795 Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13050    Accepted Submission(s): 5651 Problem Description At the entrance to the un…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 19225    Accepted Submission(s): 8042 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

题目大意:有一个h*w的公告榜,可以依次在上面添加信息.每个信息的长度为x,高为1. 优先在最上面加入,如果空间足够的话,然后优先放在最左面.统计每条公告最终的位置,即它所在的行数. 这里是线段树来存储 当前区间(i,j)的所有位置,剩余的最大空间. 初始化即为w,公告榜的宽. Problem Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h is…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 15625    Accepted Submission(s): 6580 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14337    Accepted Submission(s): 6148 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

给出一个高为h,宽为w的广告板,有n张广告需要贴,从第一行开始贴,尽量靠左,输出每个广告最后贴在哪一行的 先一直想不通这样建树是为什么 后来看到一篇题解里面的一句话“直到找到一个满足条件的叶子节点” 所以用min(h,n)建树,最后输出的为哪一行,即为一个单点(线段树的最末一层) 大概像这样 然后就是更新值,查询 #include <cstdio> #include <ctime> #include <cstring> #include <cstdlib>…

Billboard 通过这题,我知道了要活用线段树的思想,而不是拘泥于形式, 就比如这题 显然更新和查询放在一起很简单 但如果分开写 那么我觉得难度会大大增加 [题目链接]Billboard [题目类型]区间求最大值的位置update的操作在query里做了 &题意: h*w的木板,放进一些1*L的物品,求每次放空间能容纳且最上边的位子 &题解: 思路:每次找到最大值的位子,然后减去L 线段树功能:query 区间求最大值的位置(直接把update的操作在query里做了) [时间复杂度]…

1.HDU 5877  Weak Pair 2.总结:有多种做法,这里写了dfs+线段树(或+树状树组),还可用主席树或平衡树,但还不会这两个 3.思路:利用dfs遍历子节点,同时对于每个子节点au,查询它有多少个祖先av满足av<=k/au. (1)dfs+线段树 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm>…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10594    Accepted Submission(s): 4686 Problem Description At the entrance to the university, there is a huge rectangular billboard of s…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3308 题目很好懂,就是单点更新,然后求区间的最长上升子序列. 线段树区间合并问题,注意合并的条件是a[mid + 1] > a[mid],写的细心点就好了. #include <iostream> #include <cstring> #include <cstdio> using namespace std; ; struct SegTree { int l , r…

Conturbatio Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5480 Description There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place. There are also many quer…

http://acm.hdu.edu.cn/showproblem.php?pid=1828 Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2135    Accepted Submission(s): 1134 Problem Description A number of rectangular posters,…

Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16755    Accepted Submission(s): 7089 Problem Description At the entrance to the university, there is a huge rectangular billboard of…

题意: 给出一个有n个数的数列,并定义mex(l, r)表示数列中第l个元素到第r个元素中第一个没有出现的最小非负整数. 求出这个数列中所有mex的值. 思路: 可以看出对于一个数列,mex(r, r~l)是一个递增序列 mex(0, 0~n-1)是很好求的,只需要遍历找出第一个没有出现的最小非负整数就好了.这里有一个小技巧: tmp = ; ; i <= n; ++i) { mp[arr[i]] = ; while (mp.find(tmp) != mp.end()) tmp++; mex[i…