Problem:
Score : 200 points

Problem Statement

There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white.

Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i).

Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows:

If a_i = 1, he painted the region satisfying x < x_i within the rectangle.

If a_i = 2, he painted the region satisfying x > x_i within the rectangle.

If a_i = 3, he painted the region satisfying y < y_i within the rectangle.

If a_i = 4, he painted the region satisfying y > y_i within the rectangle.

Find the area of the white region within the rectangle after he finished painting.

Constraints

1 ≦ W, H ≦ 100

1 ≦ N ≦ 100

0 ≦ x_i ≦ W (1 ≦ i ≦ N)

0 ≦ y_i ≦ H (1 ≦ i ≦ N)

W, H (21:32, added), x_i and y_i are integers.

a_i (1 ≦ i ≦ N) is 1, 2, 3 or 4.

Input

The input is given from Standard Input in the following format:

W H N
x_1 y_1 a_1
x_2 y_2 a_2
:
x_N y_N a_N

Output

Print the area of the white region within the rectangle after Snuke finished painting.

Sample Input 1

5 4 2
2 1 1
3 3 4

Sample Output 1

9

The figure below shows the rectangle before Snuke starts painting.

First, as (x_1, y_1) = (2, 1) and a_1 = 1, he paints the region satisfying x < 2 within the rectangle:

Then, as (x_2, y_2) = (3, 3) and a_2 = 4, he paints the region satisfying y > 3 within the rectangle:

Now, the area of the white region within the rectangle is 9.

Sample Input 2

5 4 3
2 1 1
3 3 4
1 4 2

Sample Output 2

0

It is possible that the whole region within the rectangle is painted black.

Sample Input 3

10 10 5
1 6 1
4 1 3
6 9 4
9 4 2
3 1 3

Sample Output 3

64

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.*;
public class Main{
	public static void main(String[] args){
    	Scanner sc = new Scanner(System.in);
      	int w = sc.nextInt();
      	int h = sc.nextInt();
      	int n = sc.nextInt();
      	boolean[][] squ = new boolean[w][h];
      	squ = beTrue(squ, w, h);
      	int[] x = new int[n];
	    int[] y = new int[n];
      	int[] a = new int[n];
      	int count = 0;
      	
      	for(int i = 0; i < n; i++){
        	x[i] = sc.nextInt();
          	y[i] = sc.nextInt();
          	a[i] = sc.nextInt();
        }
      	
      	for(int i = 0; i < n; i++){
        	if(a[i]  == 1){
            	for(int j = 0; j < x[i]; j++){
                	for(int k =0; k < h; k++){
                    	squ[j][k] = false;
                    }
                }
            }else if(a[i]  == 2){
            	for(int j = x[i]; j < w; j++){
                	for(int k = 0; k < h; k++){
                    	squ[j][k] = false;
                    }
                }
            }else if(a[i]  == 3){
            	for(int j = 0; j < w; j++){
                	for(int k = 0; k < y[i]; k++){
                    	squ[j][k] = false;
                    }
                }
            }else if(a[i]  == 4){
            	for(int j = 0; j < w; j++){
                	for(int k = y[i]; k < h; k++){
                    	squ[j][k] = false;
                    }
                }
            }
        }
      	
     	for(int i = 0 ; i < w; i++){
        	for(int j = 0; j < h; j++){
            	if(squ[i][j] == true) count++;
            }
        }
      	System.out.println(count);
    }
  
  	public static boolean[][] beTrue(boolean[][] squ, int w, int h){
    	for(int i = 0; i < w; i++){
        	for(int j = 0; j < h; j++){
            	squ[i][j] = true;
            }
        }
      	return squ;
    }
}