USACO 5.3 Window Area
Window Area
IV Balkan Olympiad
You've just be assigned the project of implemented a windowing interface. This windowing interface is fairly simple, and fortunately, you don't have to display the actual windows. There are 5 basic operations:
- Create a window
- Bring a window to the top
- Put a window to the bottom
- Destroy a window
- Output what percentage of a window is visible (i.e., isn't covered by windows above it).
In the input, the operations appear in the following format:
- Create window: w(I,x,y,X,Y)
- Bring window to top: t(I)
- Put window on bottom: b(I)
- Destroy window: d(I)
- Output percentage visible: s(I)
The I is a unique identifier for each window, which is one character. The character can be any of 'a'..'z', 'A'..'Z', and '0'..'9'. No extra spaces will appear in the input.
(x,y) and (X,Y) are opposite corners of the window. When a window is created, it is put `on top'. You can't create a window with an identifier that is already in use, but you can destroy a window and then create a new one with the identifier of the destroyed window. Coordinates will be positive integers, and all windows will be of non-zero area (x != X and y != Y). The x and y coordinates are between 1 and 32767 inclusive.
PROGRAM NAME: window
INPUT FORMAT
The input file consists of a sequence of commands to your interpreter. They will be listed one per line. Terminate the program when no more input is available
SAMPLE INPUT (file window.in)
w(a,10,132,20,12)
w(b,8,76,124,15)
s(a)
OUTPUT FORMAT
Output lines only for the s() commands. Of course, there might be several s() commands (but no more than 500) so the output should be a sequence of percentages, one per line, stating the percentage of the windows that are visible. The percentages should be rounded to 3 decimal places.
SAMPLE OUTPUT (file window.out)
49.167
——————————————————————————————题解
借用USACO的字符画一下
*----------------*
| |
| |
| |
| *----* |
| | | |
| | | |
| | | |
| *----* |
| |
| |
*----------------*
||
\/
*-----*----*-----*
| | | |
| | 2 | |
| | | |
| *----* |
| | | |
| 1 | | 3 |
| | | |
| *----* |
| | 4 | |
| | | |
*-----*----*-----* *-----*
| 2 |
| |
*------------------*
| |
| |
*------------------*
| 4 |
| |
*-----* *------------*
| |
*-------| |
| | |
| | |
| 1 *------------*
| | 4 |
*-------*---* 这样上下左右的切割,递归处理
前四个操作只要记录一下高度值,修改高度就可以了
/*
ID: ivorysi
LANG: C++
PROG: window
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <set>
#include <vector>
#include <string.h>
#include <cmath>
#define siji(i,x,y) for(int i=(x);i<=(y);++i)
#define gongzi(j,x,y) for(int j=(x);j>=(y);--j)
#define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)
#define sigongzi(j,x,y) for(int j=(x);j>(y);--j)
#define inf 0x3f3f3f3f
#define ivorysi
#define mo 97797977
#define hash 974711
#define base 47
#define pss pair<string,string>
#define MAXN 5000
#define fi first
#define se second
#define pii pair<int,int>
#define esp 1e-8
typedef long long ll;
using namespace std;
char a[];
int f,cut[];
int cnt;
struct data{
int lx,ly,rx,ry;
int h;
}window[];
int calc(char c) {
if(c>='a' && c<='z') return c-'a'+;
else if (c>='A' && c<='Z') return c-'A'+;
else if(c>='' && c<='') return c-''+;
}
void ins(int x,int y,int X,int Y,char c) {
int t=calc(c);
++cnt;
window[t].lx=min(x,X);
window[t].ly=min(y,Y);
window[t].rx=max(x,X);
window[t].ry=max(y,Y);
window[t].h=;
siji(i,,) {//这里原来打成xiaosiji忘改了
if(window[i].h!= && i!=t) ++window[i].h;
}
}
void top(char c) {
int t=calc(c);
int he=window[t].h;
window[t].h=;
siji(i,,) {
if(window[i].h!= && window[i].h<he && i!=t) ++window[i].h;
}
}
void del(char c) {
int t=calc(c);
int he=window[t].h;
window[t].h=;
siji(i,,) {
if(window[i].h!= && window[i].h>he) --window[i].h;
}
}
void bottom(char c) {
int t=calc(c);
int he=window[t].h;
window[t].h=cnt;
siji(i,,) {
if(window[i].h!= && window[i].h>he && i!=t) --window[i].h;
}
}
int solute(int k,int x1,int y1,int x2,int y2) {
if(x2<=x1 || y2<=y1) return ;
int t=k;
while(t>&&(window[cut[t]].rx <=x1 || window[cut[t]].ry<=y1
|| window[cut[t]].lx>=x2 || window[cut[t]].ly>=y2)) --t;
if(t<=) return (x2-x1)*(y2-y1); int res=;
if(window[cut[t]].lx>x1) {
res+=solute(t-,x1,y1,window[cut[t]].lx,y2);
}
if(window[cut[t]].ly>y1) {
res+=solute(t-,max(x1,window[cut[t]].lx),y1,min(x2,window[cut[t]].rx),window[cut[t]].ly);
}
if(window[cut[t]].rx<x2) {
res+=solute(t-,window[cut[t]].rx,y1,x2,y2);
}
if(window[cut[t]].ry<y2) {
res+=solute(t-,max(x1,window[cut[t]].lx),window[cut[t]].ry,min(x2,window[cut[t]].rx),y2);
}
return res;
}
void solve() {
int x1,y1,x2,y2;
while(scanf("%s",a+)!=EOF) {
if(a[]=='w') {
siji(i,,strlen(a+)) {
if(a[i]<'' || a[i] > '') a[i]=' ';
}
sscanf(a+,"%d%d%d%d",&x1,&y1,&x2,&y2);
ins(x1,y1,x2,y2,a[]);
}
else if(a[]=='t') {
top(a[]);
}
else if(a[]=='b') {
bottom(a[]);
}
else if(a[]=='d') {
del(a[]);
}
else {
f=;
int t=calc(a[]);
if(window[t].h==) {puts("0.000");continue;} siji(i,,) {
if(window[t].h>window[i].h && window[i].h!=) cut[++f]=i;
}
int sa=solute(f,window[t].lx,window[t].ly,window[t].rx,window[t].ry);
int sb=(window[t].ry-window[t].ly)*(window[t].rx-window[t].lx);
printf("%.3lf\n",(double)sa/sb*);
}
}
}
int main(int argc, char const *argv[])
{
#ifdef ivorysi
freopen("window.in","r",stdin);
freopen("window.out","w",stdout);
#else
freopen("f1.in","r",stdin);
#endif
solve();
return ;
}
USACO 5.3 Window Area的更多相关文章
- [洛谷P2745] [USACO5.3]窗体面积Window Area
洛谷题目链接:[USACO5.3]窗体面积Window Area 题目描述 你刚刚接手一项窗体界面工程.窗体界面还算简单,而且幸运的是,你不必显示实际的窗体.有 5 种基本操作: 创建一个新窗体 将窗 ...
- luogu【P2745】[USACO5.3]窗体面积Window Area
这个题 就是个工程题 (然而一开始我并不知道怎么做..还是看nocow的..qwq)(原题入口) 算法为 离散化 + 扫描线 将大坐标变小 并且 用横纵坐标进行扫描 来计算面积 一开始 我想边添加 ...
- USACO 5.3 章节
相关讲解可在USACO上看原文,也可以搜索nocow找到翻译的! (nocow上有些微翻译是有问题的,如果想看nocow翻译的建议也对着英文看) 以下记录以下 自己之前未掌握的一些要点,以及按自己的括 ...
- USACO 完结的一些感想
其实日期没有那么近啦……只是我偶尔还点进去造成的,导致我没有每一章刷完的纪念日了 但是全刷完是今天啦 讲真,题很锻炼思维能力,USACO保持着一贯猎奇的题目描述,以及尽量不用高级算法就完成的题解……例 ...
- 转:SDL2源代码分析
1:初始化(SDL_Init()) SDL简介 有关SDL的简介在<最简单的视音频播放示例7:SDL2播放RGB/YUV>以及<最简单的视音频播放示例9:SDL2播放PCM>中 ...
- Using SetWindowRgn
Using SetWindowRgn Home Back To Tips Page Introduction There are lots of interesting reasons for cre ...
- SDL2源码分析2:窗体(SDL_Window)
===================================================== SDL源码分析系列文章列表: SDL2源码分析1:初始化(SDL_Init()) SDL2源 ...
- FFmpeg源代码简单分析:libavdevice的gdigrab
===================================================== FFmpeg的库函数源代码分析文章列表: [架构图] FFmpeg源代码结构图 - 解码 F ...
- SDL2源代码分析2:窗口(SDL_Window)
===================================================== SDL源代码分析系列文章列表: SDL2源代码分析1:初始化(SDL_Init()) SDL ...
随机推荐
- Hadoop基础-SequenceFile的压缩编解码器
Hadoop基础-SequenceFile的压缩编解码器 作者:尹正杰 版权声明:原创作品,谢绝转载!否则将追究法律责任. 一.Hadoop压缩简介 1>.文件压缩的好处 第一:较少存储文件占用 ...
- 转:EasyJSWebView
EasyJSWebView 是类似 Android javascriptInterface 的 uiwebview js 调用原生代码框架 示例代码: 先建一个MyJSInterface接口 @in ...
- 转:NSString / NSData / char* 类型之间的转换
1. NSString转化为UNICODE String: (NSString*)fname = @“Test”; char fnameStr[10]; memcpy(fnameStr, [fname ...
- oracle表结构和数据导出时的一些勾选项说明
使用pl/sql developer导出oracle数据库的表结构和表数据时,有一些勾选项供用户选择,需要用户根据实际情况进行勾选或取消. 导出方法如下:一.只导出表结构1.使用pl/sql deve ...
- CSS3实战之多列
CSS2中如果要设计多列布局,常用的方法有浮动和定位,但是浮动容易错位,定位无法满足模块的自适应能力,以及模块之间的文档流联动的需要.为了解决多列布局的难题,CSS3新增了多列自动布局功能. 利用多列 ...
- serialize()传值缺失
思路:serialize()获取的是 " & " 拼接的字符串,无法传值,需要拆分后,拼接,生成新字符串,传过去. 例子: var v_idd = $("form ...
- 【leetcode 简单】 第六十四题 翻转二叉树
翻转一棵二叉树. 示例: 输入: 4 / \ 2 7 / \ / \ 1 3 6 9 输出: 4 / \ 7 2 / \ / \ 9 6 3 1 备注: 这个问题是受到 Max Howell的 原问题 ...
- JavaScript的基本介绍
JavaScript入门介绍 输出语句:document.write() 1.执行顺序:从上到下,每一天语句是要加分号的,如果不加的话,浏览器会默认帮你自动添加,分号. 2.注释:一行注释就是 ...
- Python中的and和or
引子: 出现以上情况的原因是什么呢? print(bool('')) # False print(bool(0)) # False 所有变量的位操作都是通过强制转换成bool实现的,并且表达式的值是从 ...
- [转]google gflags 库完全使用
简单介绍 gflags 是 google 开源的用于处理命令行参数的项目. 安装编译 项目主页:gflags ➜ ~ git clone https://github.com/gflags/gflag ...