POJ1656 Counting Black (二维树状数组)

Counting Black
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 9655   Accepted: 6225

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:

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

Source

 
 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 
 5 using namespace std;
 6 
 7 int a[105][105];
 8 int col[105][105];
 9 
10 int lowbit(int i)
11 {
12     return i & (-i);
13 }
14 
15 void add(int x, int y, int color)
16 {
17     if(col[x][y] == color)
18         return;
19     col[x][y] = color;
20     for(int i = x; i < 105; i += lowbit(i))
21         for(int j = y; j < 105; j += lowbit(j))
22             a[i][j] += color;
23 }
24 
25 int getSum(int x, int y)
26 {
27     int ans = 0;
28     for(int i = x; i > 0; i -= lowbit(i))
29         for(int j = y; j > 0; j -= lowbit(j))
30             ans += a[i][j];
31     return ans;
32 }
33 
34 int main()
35 {
36     int T, x, y, len;
37     char cmd[10];
38     scanf("%d", &T);
39     memset(a, 0, sizeof(a));
40     memset(col, -1, sizeof(col));
41     while(T--)
42     {
43         scanf("%s%d%d%d", cmd, &x, &y, &len);
44         if(cmd[0] == 'B')
45         {
46             for(int i = x; i < x+len; ++i)
47                 for(int j = y; j < y+len; ++j)
48                     add(i, j, 1);
49         }
50         else if(cmd[0] == 'W')
51         {
52             for(int i = x; i < x+len; ++i)
53                 for(int j = y; j < y+len; ++j)
54                     add(i, j, -1);
55         }
56         else
57         {
58             printf("%d\n", getSum(x+len-1,y+len-1) + getSum(x-1,y-1) - getSum(x-1,y+len-1) - getSum(x+len-1,y-1));
59         }
60     }
61     return 0;
62 }
原文地址:https://www.cnblogs.com/dongsheng/p/3091860.html