POJ 1656 Counting Black
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 10509 | Accepted: 6786 |
Description
There is a board with 100 * 100 grids as shown below. The left-top gird is denoted as (1, 1) and the right-bottom grid is (100, 100).
![](http://poj.org/images/1656_1.jpg)
We may apply three commands to the board:
In the beginning, all the grids on the board are white. We apply a series of commands to the board. Your task is to write a program to give the numbers of black grids within a required region when a TEST command is applied.
![](http://poj.org/images/1656_1.jpg)
We may apply three commands to the board:
1. WHITE x, y, L // Paint a white square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
2. BLACK x, y, L // Paint a black square on the board,
// the square is defined by left-top grid (x, y)
// and right-bottom grid (x+L-1, y+L-1)
3. TEST x, y, L // Ask for the number of black grids
// in the square (x, y)- (x+L-1, y+L-1)
In the beginning, all the grids on the board are white. We apply a series of commands to the board. Your task is to write a program to give the numbers of black grids within a required region when a TEST command is applied.
Input
The first line of the input is an integer t (1 <= t <= 100), representing the number of commands. In each of the following lines, there is a command. Assume all the commands are legal which means that they won't try to paint/test the grids outside the board.
Output
For each TEST command, print a line with the number of black grids in the required region.
Sample Input
5
BLACK 1 1 2
BLACK 2 2 2
TEST 1 1 3
WHITE 2 1 1
TEST 1 1 3
Sample Output
7
6
CODE:
#include <iostream> #include <cstdio> #include <cstring> #define REP(i, s, n) for(int i = s; i <= n; i ++) #define REP_(i, s, n) for(int i = n; i >= s; i --) using namespace std; int main(){ int T; scanf("%d", &T); int map[100 + 10][100 + 10]; memset(map, 0, sizeof(map)); char s[7]; while(T --){ scanf("%s", s); int x, y, l; scanf("%d%d%d", &x, &y, &l); if(s[0] == 'B'){ REP(i, x, x + l - 1) REP(j, y, y + l - 1) map[i][j] = 1; } else if(s[0] == 'W'){ REP(i, x, x + l - 1) REP(j, y, y + l - 1) map[i][j] = 2; } else if(s[0] == 'T'){ int ans = 0; REP(i, x, x + l - 1) REP(j, y, y + l - 1) if(map[i][j] == 1) ans ++; printf("%d\n", ans); } } return 0; }