209 lines
6.5 KiB
Text
209 lines
6.5 KiB
Text
---
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id: rect-geo
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title: "Rectangle Geometry"
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author: Darren Yao, Michael Cao, Benjamin Qi
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description: "Geometry problems on USACO Bronze are usually quite simple and limited to intersections and unions of squares or rectangles."
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frequency: 2
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---
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import { Problem } from "../models";
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export const metadata = {
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problems: {
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blocked: [
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new Problem("Bronze", "Blocked Billboard", "759", "Easy", false, ["rect"]),
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],
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general: [
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new Problem("Bronze", "Square Pasture", "663", "Intro", false, ["rect"]),
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new Problem("Bronze", "Blocked Billboard II", "783", "Easy", false, ["rect"]),
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new Problem("CF", "Div. 3 C - White Sheet", "contest/1216/problem/C", "Normal", false, ["rect"],"See this code (TODO; codeforces is down) for a nice implementation using the Java Rectangle class."),
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]
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}
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};
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Most only include two or three squares or rectangles, in which case you can simply draw out cases on paper. This should logically lead to a solution.
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## Example: Blocked Billboard
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<problems-list problems={metadata.problems.blocked} />
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### Naive Solution
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Since all coordinates are in the range $[-1000,1000]$, we can simply go through each of the $2000^2$ possible visible squares and check which ones are visible using nested for loops.
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<spoiler title="Nested Loops">
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```cpp
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#include <bits/stdc++.h>
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using namespace std;
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bool ok[2000][2000];
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int main() {
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freopen("billboard.in","r",stdin);
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freopen("billboard.out","w",stdout);
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for (int i = 0; i < 3; ++i) {
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int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
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x1 += 1000, y1 += 1000, x2 += 1000, y2 += 1000;
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for (int x = x1; x < x2; ++x)
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for (int y = y1; y < y2; ++y) {
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if (i < 2) ok[x][y] = 1;
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else ok[x][y] = 0;
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}
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}
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int ans = 0;
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for (int x = 0; x < 2000; ++x)
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for (int y = 0; y < 2000; ++y)
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ans += ok[x][y];
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cout << ans << "\n";
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}
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```
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</spoiler>
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Of course, this wouldn't suffice if the coordinates were up to $10^9$.
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### Rectangle Class (Java)
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A useful class in `Java` for dealing with rectangle geometry problems (though probably overkill) is the built-in [`Rectangle`](https://docs.oracle.com/javase/8/docs/api/java/awt/Rectangle.html) class. To create a new rectangle, use the following constructor:
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```java
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// creates a rectangle with upper-left corner at (x,y) with a specified width and height
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Rectangle newRect = new Rectangle(x, y, width, height);
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```
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The `Rectangle` class supports numerous useful methods.
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- `firstRect.intersects(secondRect)` checks if two rectangles intersect.
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- `firstRect.union(secondRect)` returns a rectangle representing the union of two rectangles.
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- `firstRect.contains(x, y)` checks whether the integer point $(x,y)$ exists in firstRect.
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- `firstRect.intersection(secondRect)` returns a rectangle representing the intersection of two rectangles.
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- what happens when intersection is empty?
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This class can often lessen the implementation needed in a lot of bronze problems and CodeForces problems.
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For example, here is a nice implementation of the problem ([editorial](http://www.usaco.org/current/data/sol_billboard_bronze_dec17.html)).
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<spoiler title="Java Solution With Built-in Rectangle Class">
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```java
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import java.awt.Rectangle; //needed to use Rectangle class
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import java.io.*;
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import java.util.*;
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public class blockedBillboard{
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public static void main(String[] args) throws IOException{
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Scanner sc = new Scanner(new File("billboard.in"));
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PrintWriter pw = new PrintWriter(new FileWriter("billboard.out"));
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int x1, y1, x2, y2;
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//the top left point is (0,0), so you need to do -y2
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x1 = sc.nextInt(); y1 = sc.nextInt(); x2 = sc.nextInt(); y2 = sc.nextInt();
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Rectangle firstRect = new Rectangle(x1, -y2, x2-x1, y2-y1);
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x1 = sc.nextInt(); y1 = sc.nextInt(); x2 = sc.nextInt(); y2 = sc.nextInt();
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Rectangle secondRect = new Rectangle(x1, -y2, x2-x1, y2-y1);
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x1 = sc.nextInt(); y1 = sc.nextInt(); x2 = sc.nextInt(); y2 = sc.nextInt();
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Rectangle truck = new Rectangle(x1, -y2, x2-x1, y2-y1);
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long firstIntersect = getArea(firstRect.intersection(truck));
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long secondIntersect = getArea(secondRect.intersection(truck));
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pw.println(getArea(firstRect) + getArea(secondRect)
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- firstIntersect - secondIntersect);
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pw.close();
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}
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public static long getArea(Rectangle r){
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if(r.getWidth() <= 0 || r.getHeight() <= 0){
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return 0;
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}
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return (long)r.getHeight() * (long)r.getWidth();
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}
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}
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```
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</spoiler>
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<spoiler title="Java Solution Without Built-in Rectangle Class">
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```java
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import java.io.*;
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import java.util.*;
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class blockedBillboard{
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public static void main(String[] args) throws IOException{
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Scanner sc = new Scanner(new File("billboard.in"));
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PrintWriter pw = new PrintWriter(new FileWriter("billboard.out"));
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Rect a = new Rect();
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Rect b = new Rect();
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Rect t = new Rect();
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int x1, y1, x2, y2;
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a.x1 = sc.nextInt(); a.y1 = sc.nextInt(); a.x2 = sc.nextInt(); a.y2 = sc.nextInt();
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b.x1 = sc.nextInt(); b.y1 = sc.nextInt(); b.x2 = sc.nextInt(); b.y2 = sc.nextInt();
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t.x1 = sc.nextInt(); t.y1 = sc.nextInt(); t.x2 = sc.nextInt(); t.y2 = sc.nextInt();
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pw.println(area(a) + area(b) - intersect(a, t) - intersect(b, t));
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pw.close();
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sc.close();
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}
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static int area(Rect r){
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return (r.y2 - r.y1) * (r.x2 - r.x1);
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}
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static int intersect(Rect p, Rect q) {
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int xOverlap = Math.max(0, Math.min(p.x2, q.x2) - Math.max(p.x1, q.x1));
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int yOverlap = Math.max(0, Math.min(p.y2, q.y2) - Math.max(p.y1, q.y1));
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return xOverlap * yOverlap;
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}
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}
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class Rect{
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int x1, y1, x2, y2;
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Rect(){
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}
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}
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```
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</spoiler>
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### Rectangle Class (C++)
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Unfortunately, C++ doesn't have a built in rectangle class, so you need to write the functions yourself. Here is the solution to Blocked Billboard written in C++ (thanks, Brian Dean!).
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<spoiler title="C++ Solution">
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```cpp
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#include <iostream>
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#include <fstream>
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using namespace std;
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struct Rect{
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int x1, y1, x2, y2;
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};
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int area(Rect r){
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return (r.y2 - r.y1) * (r.x2 - r.x1);
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}
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int intersect(Rect p, Rect q){
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int xOverlap = max(0, min(p.x2, q.x2) - max(p.x1, q.x1));
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int yOverlap = max(0, min(p.y2, q.y2) - max(p.y1, q.y1));
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return xOverlap * yOverlap;
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}
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int main(){
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ifstream cin ("billboard.in");
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ofstream cout ("billboard.out");
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Rect a, b, t; // billboards a, b, and the truck
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cin >> a.x1 >> a.y1 >> a.x2 >> a.y2;
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cin >> b.x1 >> b.y1 >> b.x2 >> b.y2;
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cin >> t.x1 >> t.y1 >> t.x2 >> t.y2;
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cout << area(a) + area(b) - intersect(a, t) - intersect(b, t) << endl;
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return 0;
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}
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```
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</spoiler>
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## Problems
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<problems-list problems={metadata.problems.general} />
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