poj 1656 Counting Black（模拟）

Counting Black

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9101		Accepted: 5879

DescriptionThere 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).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.InputThe
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.
OutputFor 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

#include <iostream>
#define MAXNUM 101
using namespace std;

int point[MAXNUM][MAXNUM];//标记格子颜色

int main()
{
int t;//t种情况
//（x y）左上坐标 （x1,y1）右下坐标 num:记录黑色数目
int x,y,x1,y1,l,num;
string s;//命令cmd
cin>>t;//命令数目
while(t--)
{
num = 0;
cin>>s>>x>>y>>l;
x1 = x+l-1;
y1 = y+l-1;
//黑色命令，把区域内的块设置为黑色：1
if(s.compare("BLACK")==0)
{
for(int i = x; i <= x1; ++i)
for(int j = y; j <= y1; ++j)
{
point[i][j] = 1;
}
}
//白色命令，把区域内的块设置为白色：0
else if(s.compare("WHITE")==0)
{
for(int i = x; i <= x1; ++i)
for(int j = y; j <= y1; ++j)
{
point[i][j] = 0;
}
}
//测试命令，输出黑色块的数目
else if(s.compare("TEST")==0)
{
for(int i = x; i <= x1; ++i)
for(int j = y; j <= y1; ++j)
{
if(point[i][j]==1)num++;
}
cout<<num<<endl;
}
}
return 0;
}