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 C (GCC 5.4.1) (judge verdict: Accepted)

// 1: x < xi
// 2: x > xi
// 3: y < yi
// 4: y > yi

#include <stdio.h>
#define BL 0
#define WH 1
int main () {
  int W, H, N, x[100], y[100], a[100];
  int wall[100][100], whiteArea = 0;
  int index, i, j;

  // read
  scanf ("%d %d %d", &W, &H, &N);
  for (index = 0; index < N; index++) {
    scanf ("%d %d %d", &x[index], &y[index], &a[index]);
  }

  // wall init
  for (i = 0; i < H; i++) {
    for (j = 0; j < W; j++) {
      wall[i][j] = WH;
    }
  }

  // draw black
  for (index = 0; index < N; index++) {
    switch (a[index]) {
      case 1:
      for (i = 0; i < H; i++) {
        for (j = 0; j < x[index]; j++){
          wall[i][j] = BL;
        }
      }
      break;

      case 2:
      for (i = 0; i < H; i++) {
        for (j = x[index]; j < W; j++){
          wall[i][j] = BL;
        }
      }
      break;

      case 3:
      for (i = 0; i < y[index]; i++) {
        for (j = 0; j < W; j++){
          wall[i][j] = BL;
        }
      }
      break;

      case 4:
      for (i = y[index]; i < H; i++) {
        for (j = 0; j < W; j++){
          wall[i][j] = BL;
        }
      }
    }
  }

  //count WH
  for (i = 0; i < H; i++){
    for (j = 0; j < W; j++){
      whiteArea += wall[i][j];
    }
  }

  /*
  // draw wall
  for (i = 0; i < H; i++) {
    for (j = 0; j < W; j++){
      if (wall[i][j] == BL) printf ("*");
      else printf (" ");
    }
    printf ("\n");
  }
  */

  // output
  printf ("%d", whiteArea);

  return 0;
}