POJ 2482 扫描线(面积覆盖最大次数)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10806 | Accepted: 2980 |
Description
These days, having parted with friends, roommates and classmates one after another, I still cannot believe the fact that after waving hands, these familiar faces will soon vanish from our life and become no more than a memory. I will move out from school tomorrow. And you are planning to fly far far away, to pursue your future and fulfill your dreams. Perhaps we will not meet each other any more if without fate and luck. So tonight, I was wandering around your dormitory building hoping to meet you there by chance. But contradictorily, your appearance must quicken my heartbeat and my clumsy tongue might be not able to belch out a word. I cannot remember how many times I have passed your dormitory building both in Zhuhai and Guangzhou, and each time aspired to see you appear in the balcony or your silhouette that cast on the window. I cannot remember how many times this idea comes to my mind: call her out to have dinner or at least a conversation. But each time, thinking of your excellence and my commonness, the predominance of timidity over courage drove me leave silently.
Graduation, means the end of life in university, the end of these glorious, romantic years. Your lovely smile which is my original incentive to work hard and this unrequited love will be both sealed as a memory in the deep of my heart and my mind. Graduation, also means a start of new life, a footprint on the way to bright prospect. I truly hope you will be happy everyday abroad and everything goes well. Meanwhile, I will try to get out from puerility and become more sophisticated. To pursue my own love and happiness here in reality will be my ideal I never desert.
Farewell, my princess!
If someday, somewhere, we have a chance to gather, even as gray-haired man and woman, at that time, I hope we can be good friends to share this memory proudly to relight the youthful and joyful emotions. If this chance never comes, I wish I were the stars in the sky and twinkling in your window, to bless you far away, as friends, to accompany you every night, sharing the sweet dreams or going through the nightmares together.
Here comes the problem: Assume the sky is a flat plane. All the stars lie on it with a location (x, y). for each star, there is a grade ranging from 1 to 100, representing its brightness, where 100 is the brightest and 1 is the weakest. The window is a rectangle whose edges are parallel to the x-axis or y-axis. Your task is to tell where I should put the window in order to maximize the sum of the brightness of the stars within the window. Note, the stars which are right on the edge of the window does not count. The window can be translated but rotation is not allowed.
Input
There are at least 1 and at most 10000 stars in the sky. 1<=W,H<=1000000, 0<=x,y<2^31.
Output
Sample Input
3 5 4
1 2 3
2 3 2
6 3 1
3 5 4
1 2 3
2 3 2
5 3 1
Sample Output
5
6
Source
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std; #define N 10005
#define ll root<<1
#define rr root<<1|1
#define mid (a[root].l+a[root].r)/2 __int64 max(__int64 x,__int64 y){return x>y?x:y;}
__int64 min(__int64 x,__int64 y){return x<y?x:y;}
__int64 abs(__int64 x,__int64 y){return x<?-x:x;} struct node{
__int64 l, r, val, maxh;
}a[N*]; struct Line{
__int64 x1, x2, y, val;
Line(){}
Line(__int64 a,__int64 b,__int64 c,__int64 d){
x1=a;
x2=b;
y=c;
val=d;
}
}line[N*]; bool cmp(Line a,Line b){
if(a.y==b.y) return a.val>b.val;
return a.y<b.y;
} struct Po__int64{
__int64 x, y, val;
}p[N]; __int64 n, w, h;
__int64 nx;
__int64 xx[N*]; __int64 b_s(__int64 key){
__int64 l=, r=nx-;
while(l<=r){
__int64 mm=(l+r)/;
if(xx[mm]==key) return mm;
else if(xx[mm]>key) r=mm-;
else if(xx[mm]<key) l=mm+;
}
} void build(__int64 l,__int64 r,__int64 root){
a[root].l=l;
a[root].r=r;
a[root].val=a[root].maxh=;
if(l==r) return;
build(l,mid,ll);
build(mid+,r,rr);
} void up(__int64 root){
if(a[root].l==a[root].r) a[root].maxh=a[root].val;
else a[root].maxh=a[root].val+max(a[ll].maxh,a[rr].maxh);
} void update(__int64 l,__int64 r,__int64 val,__int64 root){
if(l>r) return;
if(a[root].l==l&&a[root].r==r){
a[root].val+=val;
up(root);
return;
}
if(r<=a[ll].r) update(l,r,val,ll);
else if(l>=a[rr].l) update(l,r,val,rr);
else{
update(l,mid,val,ll);
update(mid+,r,val,rr);
}
up(root);
} int main()
{
__int64 i, j, k;
while(scanf("%I64d %I64d %I64d",&n,&w,&h)==){
nx=k=;
for(i=;i<n;i++){
scanf("%I64d %I64d %I64d",&p[i].x,&p[i].y,&p[i].val);
line[k++]=Line(p[i].x,p[i].x+w-,p[i].y,p[i].val);
line[k++]=Line(p[i].x,p[i].x+w-,p[i].y+h-,-p[i].val);
xx[nx++]=p[i].x;
xx[nx++]=p[i].x+w-;
}
sort(xx,xx+nx);
nx=unique(xx,xx+nx)-xx;
build(,nx,);
__int64 ans=;
sort(line,line+k,cmp);
for(i=;i<k-;i++){
update(b_s(line[i].x1),b_s(line[i].x2),line[i].val,);
ans=max(ans,a[].maxh);
}
printf("%I64d\n",ans);
}
return ;
}
POJ 2482 扫描线(面积覆盖最大次数)的更多相关文章
- poj 2482 Stars in Your Window(扫描线)
id=2482" target="_blank" style="">题目链接:poj 2482 Stars in Your Window 题目大 ...
- HDU 1255 覆盖的面积 (线段树扫描线+面积交)
自己YY了一个的写法,不过时间复杂度太高了,网上的想法太6了 题意:给你一些矩阵,求出矩阵的面积并 首先按照x轴离散化线段到线段树上(因为是找连续区间,所以段建树更加好做). 然后我们可以想一下怎样 ...
- 【POJ 2482】 Stars in Your Window(线段树+离散化+扫描线)
[POJ 2482] Stars in Your Window(线段树+离散化+扫描线) Time Limit: 1000MS Memory Limit: 65536K Total Submiss ...
- hdu 1255 覆盖的面积 (线段树处理面积覆盖问题(模板))
http://acm.hdu.edu.cn/showproblem.php?pid=1255 覆盖的面积 Time Limit: 10000/5000 MS (Java/Others) Memo ...
- POJ 2482 Stars in Your Window(线段树)
POJ 2482 Stars in Your Window 题目链接 题意:给定一些星星,每一个星星都有一个亮度.如今要用w * h的矩形去框星星,问最大能框的亮度是多少 思路:转化为扫描线的问题,每 ...
- POJ 1151 扫描线 线段树
题意:给定平面直角坐标系中的N个矩形,求它们的面积并. 题解:建立一个四元组(x,y1,y2,k).(假设y1<y2)用来储存每一条线,将每一条线按x坐标排序.记录所有的y坐标以后排序离散化.离 ...
- poj 1265 Area 面积+多边形内点数
Area Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5861 Accepted: 2612 Description ...
- n个点m条有向边,求在入度为零的点到n号点的所有路 //径中,哪条边被这些路径覆盖的次数最多
//n个点m条有向边,求在入度为零的点到n号点的所有路 //径中,哪条边被这些路径覆盖的次数最多 //有关DAG的知识,先记个模板 #include<iostream> #include& ...
- hdu1255扫描线计算覆盖两次面积
总体来说也是个模板题,但是要开两个线段树来保存被覆盖一次,两次的面积 #include<iostream> #include<cstring> #include<cstd ...
随机推荐
- openerp安装记录及postgresql数据库问题解决
ubuntu-14.04下openerp安装记录1.安装PostgreSQL 数据库 a.安装 sudo apt-get install postgresql 安装后ubu ...
- Java编程思想学习笔记_4(异常机制,容器)
一.finally语句注意的细节: 当涉及到break和continue语句的时候,finally字句也会得到执行. public class Test7 { public static void m ...
- FLAG_ACTIVITY_CLEAR_TOP和FLAG_ACTIVITY_REORDER_TO_FRONT用法
Activity的两种启动模式: FLAG_ACTIVITY_CLEAR_TOP和FLAG_ACTIVITY_REORDER_TO_FRONT 1. 如果已经启动了四个Activity:A,B,C和D ...
- Android中Activity、Service和线程之间的通信
Activity.Service和线程应该是Android编程中最常见的几种类了,几乎大多数应用程序都会涉及到这几个类的编程,自然而然的,也就会涉及到三者之间的相互通信,本文就试图简单地介绍一下这三者 ...
- golang代码执行顺序
一:首先man.go,整个程序的入口 func main() { beego.Run() } 然后beego.run()代码 // Run beego application. // beego.Ru ...
- Android 高手进阶之自定义View,自定义属性(带进度的圆形进度条)
Android 高手进阶(21) 版权声明:本文为博主原创文章,未经博主允许不得转载. 转载请注明地址:http://blog.csdn.net/xiaanming/article/detail ...
- @responseBody注解的使用
1. @responseBody注解的作用是将controller的方法返回的对象通过适当的转换器转换为指定的格式之后,写入到response对象的body区,通常用来返回JSON数据或者是XML 数 ...
- Hexo+github 搭建个人博客(一)
一.软件环境准备 1.安装git windows下载exe安装:linux 执行 apt-get install git-core 安装 2.安装Node.js windows使用 msi 文件进行安 ...
- SQL 比较时间大小
比较字符串类型的时间大小 数据库中的时间是varchar类型的,MySql使用CURDATE()来获取当前日期,SqlServer通过GETDATE()来获取当前日期 1. 直接使用字符串来比较 注意 ...
- python 集合
面向对象的集合: #coding:utf-8 __author__ = 'similarface' class Set: ''' list实现集合,及其集合操作 ''' def __init__(se ...