Luogu1502 窗口的星星 (线段树扫描线)
将每个点拓展为矩形,将\(y\)离散,延\(x\)轴扫描,每次更新最值
用了一百年的pushdown操作疑似有问题,亦或这道题特殊,我乱改了pushdown位置就过了,我能怎么办,WA了一发,y数组没开够又RE了一发。。。
话说POJ上的情书让我回忆起童年那个彪悍的女孩,一晃十年了
Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walking out of the classroom and turned your head back, with the soft sunset glow shining on your rosy cheek, I knew, I knew that I was already drunk on you. Then, after several months’ observation and prying, your grace and your wisdom, your attitude to life and your aspiration for future were all strongly impressed on my memory. You were the glamorous and sunny girl whom I always dream of to share the rest of my life with. Alas, actually you were far beyond my wildest dreams and I had no idea about how to bridge that gulf between you and me. So I schemed nothing but to wait, to wait for an appropriate opportunity. Till now — the arrival of graduation, I realize I am such an idiot that one should create the opportunity and seize it instead of just waiting.
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.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
//#define ON_DEBUG
#ifdef ON_DEBUG
#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x) cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#endif
struct ios{
template<typename ATP>ios& operator >> (ATP &x){
x = 0; int f = 1; char c;
for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
x*= f;
return *this;
}
}io;
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
const int N = 10007;
struct Line{
int X, Y1, Y2, w;
bool operator < (const Line &com) const{
if(X != com.X) return X < com.X;
return w > com.w;
}
}a[N << 1];
struct Tree{
int mx, tag;
}t[N << 3];
inline void Pushup(int rt){
t[rt].mx = Max(t[rt << 1].mx, t[rt << 1 | 1].mx);
}
inline void Pushdown(int rt, int l, int r){
if(!t[rt].tag) return;
t[rt].mx += t[rt].tag;
if(l != r){
t[rt << 1].tag += t[rt].tag;
t[rt << 1 | 1].tag += t[rt].tag;
}
t[rt].tag = 0;
}
inline void Updata(int rt, int l, int r, int L, int R, int w){
if(L <= l && r <= R){
t[rt].tag += w;
// Pushdown(rt, l, r);
return;
}
// Pushdown(rt, l, r);
int mid = (l + r) >> 1;
if(L <= mid) Updata(lson, L, R, w);
if(R > mid) Updata(rson, L, R, w);
Pushdown(lson), Pushdown(rson); // QAQ
Pushup(rt);
}
int y[N << 1];
int main(){
int Tasks;
io >> Tasks;
while(Tasks--){
Fill(t, 0);
int n, W, H;
io >> n >> W >> H;
R(i,1,n){
int X, Y, val;
io >> X >> Y >> val;
a[i] = (Line){X, Y, Y + H - 1, val};
a[i + n] = (Line){X + W - 1, Y, Y + H - 1, -val};
y[i] = Y;
y[i + n] = Y + H - 1;
}
n <<= 1;
sort(a + 1, a + n + 1);
sort(y + 1, y + n + 1);
int m = unique(y + 1, y + n + 1) - y - 1;
int ans = 0;
R(i,1,n){
int l = lower_bound(y + 1, y + m + 1, a[i].Y1) - y;
int r = lower_bound(y + 1, y + m + 1, a[i].Y2) - y;
Updata(1, 1, m, l, r, a[i].w);
ans = Max(ans, t[1].mx);
}
printf("%d\n", ans);
}
return 0;
}

破案了,我pushdown傻逼了, 一中午愉悦的聊天后 终于圆满了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long
//#define ON_DEBUG
#ifdef ON_DEBUG
#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x) cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#endif
struct ios{
template<typename ATP>ios& operator >> (ATP &x){
x = 0; int f = 1; char c;
for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
x*= f;
return *this;
}
}io;
using namespace std;
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
const int N = 10007;
struct Line{
int X, Y1, Y2, w;
bool operator < (const Line &com) const{
if(X != com.X) return X < com.X;
return w > com.w;
}
}a[N << 1];
struct Tree{
int mx, tag;
}t[N << 3];
inline void Pushup(int rt){
t[rt].mx = Max(t[rt << 1].mx, t[rt << 1 | 1].mx);
}
inline void Pushdown(int rt, int l, int r){
if(!t[rt].tag) return;
t[rt << 1].mx += t[rt].tag;
t[rt << 1 | 1].mx += t[rt].tag;
t[rt << 1].tag += t[rt].tag;
t[rt << 1 | 1].tag += t[rt].tag;
t[rt].tag = 0;
}
inline void Updata(int rt, int l, int r, int L, int R, int w){
if(L <= l && r <= R){
t[rt].mx += w;
t[rt].tag += w;
return;
}
Pushdown(rt, l, r);
int mid = (l + r) >> 1;
if(L <= mid) Updata(lson, L, R, w);
if(R > mid) Updata(rson, L, R, w);
Pushup(rt);
}
int y[N << 1];
int main(){
FileOpen();
int Tasks;
io >> Tasks;
while(Tasks--){
Fill(t, 0);
int n, W, H;
io >> n >> W >> H;
R(i,1,n){
int X, Y, val;
io >> X >> Y >> val;
a[i] = (Line){X, Y, Y + H - 1, val};
a[i + n] = (Line){X + W - 1, Y, Y + H - 1, -val};
y[i] = Y;
y[i + n] = Y + H - 1;
}
n <<= 1;
sort(a + 1, a + n + 1);
sort(y + 1, y + n + 1);
int m = unique(y + 1, y + n + 1) - y - 1;
int ans = 0;
R(i,1,n){
int l = lower_bound(y + 1, y + m + 1, a[i].Y1) - y;
int r = lower_bound(y + 1, y + m + 1, a[i].Y2) - y;
Updata(1, 1, m, l, r, a[i].w);
ans = Max(ans, t[1].mx);
}
printf("%d\n", ans);
}
return 0;
}

Luogu1502 窗口的星星 (线段树扫描线)的更多相关文章
- 51nod 1208 窗上的星星 | 线段树 扫描线
51nod 1208 Stars In Your Window 题面 整点上有N颗星星,每颗星星有一个亮度.用一个平行于x轴和y轴,宽为W高为H的方框去套星星.套住的所有星星的亮度之和为S(包括边框上 ...
- 【学习笔记】线段树—扫描线补充 (IC_QQQ)
[学习笔记]线段树-扫描线补充 (IC_QQQ) (感谢 \(IC\)_\(QQQ\) 大佬授以本内容的著作权.此人超然于世外,仅有 \(Luogu\) 账号 尚可膜拜) [学习笔记]线段树详解(全) ...
- 【Codeforces720D】Slalom 线段树 + 扫描线 (优化DP)
D. Slalom time limit per test:2 seconds memory limit per test:256 megabytes input:standard input out ...
- Codeforces VK CUP 2015 D. Closest Equals(线段树+扫描线)
题目链接:http://codeforces.com/contest/522/problem/D 题目大意: 给你一个长度为n的序列,然后有m次查询,每次查询输入一个区间[li,lj],对于每一个查 ...
- 【POJ-2482】Stars in your window 线段树 + 扫描线
Stars in Your Window Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11706 Accepted: ...
- HDU 4419 Colourful Rectangle --离散化+线段树扫描线
题意: 有三种颜色的矩形n个,不同颜色的矩形重叠会生成不同的颜色,总共有R,G,B,RG,RB,GB,RGB 7种颜色,问7种颜色每种颜色的面积. 解法: 很容易想到线段树扫描线求矩形面积并,但是如何 ...
- BZOJ-3228 棋盘控制 线段树+扫描线+鬼畜毒瘤
3228: [Sdoi2008]棋盘控制 Time Limit: 10 Sec Memory Limit: 128 MB Submit: 23 Solved: 9 [Submit][Status][D ...
- BZOJ-3225 立方体覆盖 线段树+扫描线+乱搞
看数据范围像是个暴力,而且理论复杂度似乎可行,然后被卡了两个点...然后来了个乱搞的线段树+扫描线.. 3225: [Sdoi2008]立方体覆盖 Time Limit: 2 Sec Memory L ...
- hdu 5091(线段树+扫描线)
上海邀请赛的一道题目,看比赛时很多队伍水过去了,当时还想了好久却没有发现这题有什么水题的性质,原来是道成题. 最近学习了下线段树扫描线才发现确实是挺水的一道题. hdu5091 #include &l ...
- POJ1151+线段树+扫描线
/* 线段树+扫描线+离散化 求多个矩形的面积 */ #include<stdio.h> #include<string.h> #include<stdlib.h> ...
随机推荐
- 为什么我在css里使用功能类优先
前言 我想在我们开始的学CSS语法的时候,都是从以下的流程开始的: 1.写一个CSS类选择器: .my-class { } 2.往选择器里填充CSS语法: .my-class { display fl ...
- 使用 KubeKey 搭建 Kubernetes/KubeSphere 环境的"心路(累)历程"
目录 今天要干嘛? 在哪里干? 从哪里开始干? 快速开干! 解决依赖问题再继续干! 如何干翻重来? 连着 KubeSphere 一起干! 干不过,输了. 重整旗鼓,继续干! 再次重整旗鼓,继续干! 一 ...
- torch.rand、torch.randn、torch.normal、torch.linespace
torch.rand(*size, *, out=None, dtype=None, layout=torch.strided, device=None, requires_grad=False) # ...
- 【lora无线数传通信模块】亿佰特E22串口模块用于物联网地震预警传感通信方案
物联网地震预警项目介绍: 地震,俗称地动.它像平常的刮风下雨一样,是一种常见的自然现象,是地壳运动的一种表现,即地球内部缓慢积累的能量突然释放而引起的地球表层的振动.据统计,5级以上地震就能够造成破坏 ...
- SpringBoot整合RabbitMQ实战附加死信交换机
前言 使用springboot,实现以下功能,有两个队列1.2,往里面发送消息,如果处理失败发生异常,可以重试3次,重试3次均失败,那么就将消息发送到死信队列进行统一处理,例如记录数据库.报警等 环境 ...
- Acwing787.归并排序
Acwing787.归并排序 归并模板 归并排序,合二为一 题目链接:Acwing787.归并排序 #include<iostream> using namespace std; cons ...
- Python中的逻辑表达式
首先要明确一点,Python的逻辑运算符,可以用来操作任何类型的表达式(不局限于Bool类型),且运算后的结果也不一定是Bool类型的,而是其左右其中一个表达式的值 表达式1 and 表达式2 pyt ...
- SAP 定义用户组
SUGR,可进行创建.查看.删除等维护性操作,并可指定本组的用户
- XAMPP Apache安装时问题总结
昨天遇到的一个问题,安装XAMPP后无法启动Apache服务,本以为可能是一些系统服务文件被占用的原因,安装软件后需重启一遍.可是重启计算机后依然无法启动Apache服务,状态栏里总是显示" ...
- Python爬虫常用:谷歌浏览器驱动——Chromedriver 插件安装教程
我们在做爬虫的时候经常要使用谷歌浏览器驱动,今天分享下这个Chromedriver 插件的安装方法. 第一步:打开谷歌浏览器打开设置面板 嫌枯燥的小伙伴可以点击此处找管理员小姐姐领取免费资料 第二步: ...