POJ-2779-Counting Black

一、题目

2779:Counting Black

• http://poj.grids.cn/practice/2779/

时间限制: 

1000ms

 

内存限制: 

65536kB

描述

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). 

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.

输入

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.

输出

For each TEST command, print a line with the number of black grids in the required region.

样例输入

5
BLACK 1 1 2
BLACK 2 2 2
TEST 1 1 3
WHITE 2 1 1
TEST 1 1 3

样例输出

7
6

二、分析

    建立数组按指令模拟操作　　

三、AC源代码

#include <stdio.h>
int main()
{
    int n,i,j,x,y,l,count;
    char color[6];
    int grid[100][100]={0};
    scanf("%d",&n);
    while(n--)
    {
        count=0;
        scanf("%s%d%d%d",color,&x,&y,&l);
        for(i=x;i<x+l;i++)
            for(j=y;j<y+l;j++)
                if(color[0]=='B')
                    grid[i][j]=1;
                else if (color[0]=='W')
                    grid[i][j]=0;
                else if(grid[i][j]==1)
                    count++;
        if(color[0]=='T')
            printf("%d\n",count);
    }
    return 0;
}