POJ 1656 二维树状数组简单应用

一个水题。。。结果因为一个加号写成了减号。。。硬是耗费了一个多小时。。。T^T

Counting Black

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10012		Accepted: 6466

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

就是多次统计某一个区域里黑色方块的数量，有人说用直接模拟就能过去。。。我没试。。。为了保险还是用树状数组撸了。。。

用一个atlas数组来记录某一个状态时黑白方块的数量（奇怪的是。。我用bool数组的话 time是16ms。。。如果是int数组的话 time是0ms。。。）atlas[i][j]为1表示black，为0表示white。

然后一次一次查询时修改即可。

至于二维树状数组的写法，只要在一维的基础上加一层循环即可。

下面放出代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>

using namespace std;

#define MAX_N 101

int bintree[MAX_N][MAX_N] = { 0 };
int atlas[MAX_N][MAX_N] = { 0 };

int lowbit(int pos) {
return pos & (-pos);
}

int getsum(int xpos, int ypos) {
int sum = 0;
while(xpos > 0) {
int yypos = ypos;
while(yypos > 0)
sum += bintree[xpos][yypos], yypos-=lowbit(yypos);
xpos -= lowbit(xpos);
}
return sum;
}

void update(int xpos, int ypos, int val) {
while(xpos <= 100) {
int yypos = ypos;
while(yypos <= 100)
bintree[xpos][yypos] += val, yypos += lowbit(yypos);
xpos += lowbit(xpos);
}
}

int main() {
int test_n, x, y, l;
string order;
cin >> test_n;
while(test_n--) {
cin >> order >> x >> y >> l;
if(order == "BLACK") {
for(int i = x; i <= x + l - 1; i++) {
for(int j = y; j <= y + l - 1; j++) {
if(atlas[i][j] == 0)
update(i, j, 1), atlas[i][j] = 1;
}
}
}
else if(order == "WHITE") {
for(int i = x; i <= x + l - 1; i++) {
for(int j = y; j <= y + l - 1; j++) {
if(atlas[i][j] == 1)
update(i, j, -1), atlas[i][j] = 0;
}
}
}
else
printf("%d\n", getsum(x + l - 1, y + l - 1) + getsum(x - 1, y - 1) - getsum(x - 1, y + l - 1) - getsum(x + l - 1, y - 1));
}
return 0;
}