82 lines
3.3 KiB
Java
82 lines
3.3 KiB
Java
import java.io.*;
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import java.util.*;
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public class billboard {
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public static void main(String[] args) throws IOException {
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// initialize file I/O
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BufferedReader br = new BufferedReader(new FileReader("billboard.in"));
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PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("billboard.out")));
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// read in the locations of the first billboard
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StringTokenizer st = new StringTokenizer(br.readLine());
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int x1 = Integer.parseInt(st.nextToken());
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int y1 = Integer.parseInt(st.nextToken());
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int x2 = Integer.parseInt(st.nextToken());
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int y2 = Integer.parseInt(st.nextToken());
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// read in the locations of the second billboard
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st = new StringTokenizer(br.readLine());
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int x3 = Integer.parseInt(st.nextToken());
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int y3 = Integer.parseInt(st.nextToken());
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int x4 = Integer.parseInt(st.nextToken());
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int y4 = Integer.parseInt(st.nextToken());
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// read in the locations of the truck
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st = new StringTokenizer(br.readLine());
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int x5 = Integer.parseInt(st.nextToken());
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int y5 = Integer.parseInt(st.nextToken());
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int x6 = Integer.parseInt(st.nextToken());
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int y6 = Integer.parseInt(st.nextToken());
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br.close();
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// the visible area is the sum of the visible area of the first billboard and the second billboard
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int combinedArea = visibleArea(x1, y1, x2, y2, x5, y5, x6, y6) + visibleArea(x3, y3, x4, y4, x5, y5, x6, y6);
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// print the answer
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pw.println(combinedArea);
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pw.close();
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}
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/**
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* Given the lower-left and upper-right corners of a rectangle, return the area of the rectangle
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* @param x1 x-coordinate of lower-left corner
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* @param y1 y-coordinate of lower-left corner
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* @param x2 x-coordinate of upper-right corner
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* @param y2 y-coordinate of upper-right corner
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* @return area of the rectangle
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*/
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public static int areaOfRectangle(int x1, int y1, int x2, int y2) {
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return (x2-x1)*(y2-y1);
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}
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/**
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* Given the corners of two rectangles, return the area inside the first rectangle
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* but outside the second
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* @param x1 x-coordinate of lower-left corner of first rectangle
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* @param y1 y-coordinate of lower-left corner of first rectangle
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* @param x2 x-coordinate of upper-right corner of first rectangle
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* @param y2 y-coordinate of upper-right corner of first rectangle
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* @param x3 x-coordinate of lower-left corner of second rectangle
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* @param y3 y-coordinate of upper-right corner of second rectangle
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* @param x4 x-coordinate of lower-left corner of second rectangle
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* @param y4 y-coordinate of upper-right corner of second rectangle
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* @return
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*/
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public static int visibleArea(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) {
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// start by computing the area that would be visible if there were no second rectangle
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int visibleArea = areaOfRectangle(x1, y1, x2, y2);
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// compute the boundaries of the intersection
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int leftmostBlockedX = Math.max(x1, x3);
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int rightmostBlockedX = Math.min(x2, x4);
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int bottommostBlockedY = Math.max(y1, y3);
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int topmostBlockedY = Math.min(y2, y4);
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// if the second rectangle does exist, subtract out the area that it blocks
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if(leftmostBlockedX < rightmostBlockedX && bottommostBlockedY < topmostBlockedY) {
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visibleArea -= areaOfRectangle(leftmostBlockedX, bottommostBlockedY, rightmostBlockedX, topmostBlockedY);
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}
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return visibleArea;
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}
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} |