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 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 the dining room menu, and other important information.

On September 1, the billboard was empty. One by one, the announcements started being put on the billboard.

Each announcement is a stripe of paper of unit height. More specifically, the i-th announcement is a rectangle of size 1 * wi.

When someone puts a new announcement on the billboard, she would always choose the topmost possible position for the announcement. Among all possible topmost positions she would always choose the leftmost one.

If there is no valid location for a new announcement, it is not put on the billboard (that's why some programming contests have no participants from this university).

Given the sizes of the billboard and the announcements, your task is to find the numbers of rows in which the announcements are placed.

 
Input
There are multiple cases (no more than 40 cases).

The first line of the input file contains three integer numbers, h, w, and n (1 <= h,w <= 10^9; 1 <= n <= 200,000) - the dimensions of the billboard and the number of announcements.

Each of the next n lines contains an integer number wi (1 <= wi <= 10^9) - the width of i-th announcement.

 
Output
For each announcement (in the order they are given in the input file) output one number - the number of the row in which this announcement is placed. Rows are numbered from 1 to h, starting with the top row. If an announcement can't be put on the billboard, output "-1" for this announcement.
 
Sample Input
3 5 5
2
4
3
3
 
Sample Output
1
2
1
3
-1
 
Author
hhanger@zju
 
Source
 
Recommend
lcy
 

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
思路:每次找到最大值的位子,然后减去L
线段树功能:query:区间求最大值的位子

 #include<cstdio>
#include<string.h>
#include<algorithm> #define clr(x,y) memset(x,y,sizeof(x))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1 const int N=2e5+;
using namespace std; int h,w,n,MAX[N<<]; void PushUp(int rt)
{
MAX[rt]=max(MAX[rt<<],MAX[rt<<|]);
} void build(int l,int r,int rt)
{
int m; MAX[rt]=w;
if(l==r) {
return;
} m=(l+r)>>;
build(lson);
build(rson);
} int query(int x,int l,int r,int rt)
{
int m,ret;
if(l==r) {
MAX[rt]-=x;
return l;
}
m=(l+r)>>;
ret=(MAX[rt<<]>=x) ? query(x,lson) : query(x,rson);
PushUp(rt);
return ret;
} int main()
{
int x;
while(~scanf("%d%d%d",&h,&w,&n)) {
if(h>n) h=n;
build(,h,);
while(n--) {
scanf("%d",&x);
if(MAX[]<x) puts("-1");
else printf("%d\n",query(x,,h,)); } } return ;
}

[HDU] 2795 Billboard [线段树区间求最值]的更多相关文章

  1. HUD.2795 Billboard ( 线段树 区间最值 单点更新 单点查询 建树技巧)

    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 ...

  2. hdu 1754 I Hate It(线段树区间求最值)

    I Hate It Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total S ...

  3. hdu4521-小明系列问题——小明序列(线段树区间求最值)

    题意:求最长上升序列的长度(LIS),但是要求相邻的两个数距离至少为d,数据范围较大,普通dp肯定TLE.线段树搞之就可以了,或者优化后的nlogn的dp. 代码为  线段树解法. #include ...

  4. ACM学习历程—HDU 2795 Billboard(线段树)

    Description At the entrance to the university, there is a huge rectangular billboard of size h*w (h ...

  5. poj3264(线段树区间求最值)

    题目连接:http://poj.org/problem?id=3264 题意:给定Q(1<=Q<=200000)个数A1,A2,```,AQ,多次求任一区间Ai-Aj中最大数和最小数的差. ...

  6. HDU 2795 Billboard (线段树单点更新 && 求区间最值位置)

    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 ...

  7. HDU 2795 Billboard 线段树,区间最大值,单点更新

    Billboard Time Limit: 20000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

  8. HDU 2795.Billboard-完全版线段树(区间求最值的位置、区间染色、贴海报)

    HDU2795.Billboard 这个题的意思就是在一块h*w的板子上贴公告,公告的规格为1*wi ,张贴的时候尽量往上,同一高度尽量靠左,求第n个公告贴的位置所在的行数,如果没有合适的位置贴则输出 ...

  9. HDU 2795 Billboard (线段树+贪心)

    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 ...

随机推荐

  1. Android开发的体会

    View Functionality--------------->逻辑 Data----------------------->实体,包括从View.Net.其他对象比如Location ...

  2. 关于popupwindow的两种实现方式

    http://104zz.iteye.com/blog/1685389 android PopupWindow实现从底部弹出或滑出选择菜单或窗口 本实例弹出窗口主要是继承PopupWindow类来实现 ...

  3. android手机端保存xml数据

    1.前面写的这个不能继续插入数据,今天补上,当文件不存在的时候就创建,存在就直接往里面添加数据. 2.代码如下: <pre name="code" class="j ...

  4. 【转】Android中BindService方式使用的理解

    原文网址:http://www.cnblogs.com/onlylittlegod/archive/2011/05/15/2046652.html 最近学习了一下Android里面的Service的应 ...

  5. 股票市场问题(The Stock Market Problem)

    Question: Let us suppose we have an array whose ith element gives the price of a share on the day i. ...

  6. Codeforce 216 div2

    D 只要搞清楚一个性质:确定了当前最大和次大的位置,局面就唯一确定了; 根据这个性质设计dp,统计到达该局面的方法数即可. E 询问的要求是: 求有多少个区间至少覆盖了询问的点集中的一个; 转化成逆命 ...

  7. scp 对拷文件夹 和 文件夹下的所有文件 对拷文件并重命名

    对拷文件夹 (包括文件夹本身) scp -r   /home/wwwroot/www/charts/util root@192.168.1.65:/home/wwwroot/limesurvey_ba ...

  8. UBI(unsorted block image )块管理

    一.介绍 ubi是unsorted block images的缩写,是由IBM开发设计的,它与ubifs有不同的含义,ubifs是一种文件系统(nokia开发的):而ubi是一种块管理工具,工作在mt ...

  9. 传智播客学习之HTML基础语法

    一.基本格式 1.不用区分大小写. 2.HTML代码由<html>开始</html>结束.里面由头部分<head></head>和体部分<body ...

  10. 字典树-百度之星-Xor Sum

    Xor Sum Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包括了N个正整数,随后 Prometheu ...