POJ 1656 Counting Black
Counting Black
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 9772 | Accepted: 6307 |
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).
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.
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
题目大意：输入n代表有n行指令，每行指令包括一个字符串cmd和三个数字x,y,nLen,如果cmd为“BLACK”表示将以x,y为左上角的边长为nLen的正方形里面的方块全部变成黑色，如果cmd为“WHITE”表示将以x,y为左上角的边长为nLen的正方形里面的方块全部变成白色，如果cmd为“TEST”则查询以x,y为左上角的边长为nLen的正方形里面的黑色方块共有多少个。
解题方法：而为树状数组。
#include <stdio.h> #include <iostream> #include <string.h> using namespace std; int maze[105][105]; int color[105][105]; int lowbit(int x) { return x & -x; } void add(int x, int y, int n) { if (color[x][y] == n) { return ; } color[x][y] = n; for (int i = x; i <= 100; i += lowbit(i)) { for (int j = y; j <= 100; j += lowbit(j)) { maze[i][j] += n; } } } int getsum(int x, int y) { int sum = 0; for (int i = x; i >= 1; i -= lowbit(i)) { for (int j = y; j >= 1; j -= lowbit(j)) { sum += maze[i][j]; } } return sum; } int main() { int n; int x, y, nLen; char cmd[10]; scanf("%d", &n); memset(color, -1, sizeof(color)); while(n--) { scanf("%s%d%d%d", cmd, &x, &y, &nLen); if (strcmp(cmd, "BLACK") == 0) { for (int i = x; i < x + nLen; i++) { for (int j = y; j < y + nLen; j++) { add(i, j, 1); } } } else { if (strcmp(cmd, "WHITE") == 0) { for (int i = x; i < x + nLen; i++) { for (int j = y; j < y + nLen; j++) { add(i, j, -1); } } } else { printf("%d\n", getsum(x + nLen - 1, y + nLen - 1) - getsum(x - 1, y + nLen - 1) - getsum(x + nLen - 1, y - 1) + getsum(x - 1, y - 1)); } } } return 0; }