(Analysis by Nick Wu)

In this problem, we have a rectangle that is partially covered by another rectangle. We wish to compute the smallest area axis-aligned rectangle that covers the remainder of the rectangle.

After drawing out some examples on paper, we see that there are three different cases. The rectangle might have four of its corners covered, two of its corners covered, or fewer than two of its corners covered. If it has four of its corners covered, then it must be completely covered, in which case the area is zero. If it has two of its corners covered, then we can remove just the intersection of the area of the two rectangles and the remaining area can be covered exactly. If it has fewer than two corners, we must cover the entire rectangle, since one of the pairs of opposite corners will remain uncovered.

import java.io.*;
import java.util.*;
public class billboard {
public static void main(String[] args) throws IOException {
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("billboard.out")));

// read in the locations of the corners of the rectangles
int x1 = Integer.parseInt(st.nextToken());
int y1 = Integer.parseInt(st.nextToken());
int x2 = Integer.parseInt(st.nextToken());
int y2 = Integer.parseInt(st.nextToken());

int x3 = Integer.parseInt(st.nextToken());
int y3 = Integer.parseInt(st.nextToken());
int x4 = Integer.parseInt(st.nextToken());
int y4 = Integer.parseInt(st.nextToken());

// count how many corners of the billboard are covered
int cornerCover = 0;
if(covered(x1, y1, x3, y3, x4, y4)) cornerCover++;
if(covered(x1, y2, x3, y3, x4, y4)) cornerCover++;
if(covered(x2, y1, x3, y3, x4, y4)) cornerCover++;
if(covered(x2, y2, x3, y3, x4, y4)) cornerCover++;
// if fewer than 2 corners are covered, the whole rectangle must be covered
if(cornerCover < 2) {
pw.println((x2-x1)*(y2-y1));
}
// if all 4 corners are covered, then nothing needs to be covered
else if(cornerCover == 4) {
pw.println(0);
}
else {
// we only need to cover some portion of the rectangle
// find the intersection of the two rectangles
int xL = Math.max(x1, x3);
int xR = Math.min(x2, x4);
int yL = Math.max(y1, y3);
int yR = Math.min(y2, y4);
// subtract away the area of the intersection
pw.println((x2-x1)*(y2-y1) - (xR-xL)*(yR-yL));
}
pw.close();
}

public static boolean covered(int x, int y, int x1, int y1, int x2, int y2) {
// returns true if (x, y) is covered by the rectangle bounded by (x1, y1) and (x2, y2)
// returns false otherwise
return x >= x1 && x <= x2 && y >= y1 && y <= y2;
}

}