HDU 2642 Stars（二维树状数组模板题）_二维树状数组例题-优快云博客

本文链接：https://blog.youkuaiyun.com/zzcblogs/article/details/73302972

本文介绍了一个用于计算指定区域内亮星数量的算法。该算法通过处理一系列操作指令（点亮或熄灭星星），并针对查询指令快速反馈指定区域内亮星的数量。采用二维前缀和结合差分更新的方法实现高效的数据处理。

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Stars

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)

Total Submission(s): 118 Accepted Submission(s): 56

Problem Description

Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y" where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.

There is only one case.

Input

The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.

Output

For each query,output the number of bright stars in one line.

Sample Input

5
B 581 145
B 581 145
Q 0 600 0 200
D 581 145
Q 0 600 0 200

Sample Output

1
0

没啥好说的。。。注意范围 0不可取，要都加一。

全局变量默认初始化为0不需要 memset

#include <iostream>
#include <algorithm>
#include <array>
#include <string.h>
#include <cmath>
#include <stdio.h>
using namespace std;
int flag[1005][1005];
int tree[1005][1005];
int low_bit(int x) {
    return x & (-x);
}

void update(int pos_x, int pos_y, int value) {
    for (int i = pos_x ; i <= 1003; i += low_bit(i)) {
        for (int j = pos_y; j <= 1003; j += low_bit(j)) {
            tree[i][j] += value;
        }
    }
}

int sum(int pos_x, int pos_y) {
    int sum = 0;
    for (int i = pos_x; i > 0; i -= low_bit(i)) {
        for (int j = pos_y; j > 0; j -= low_bit(j)) {
            sum += tree[i][j];
        }
    }
    return sum;
}
int main () {
    int n;
    cin >> n;
    while (n--) {
        char ch[2];
        cin >> ch;
        if (ch[0] == 'B') {
            int a, b;
            cin >> a >> b;
            a++, b++;
            if (!flag[a][b]) {
                flag[a][b] = 1;
                update(a, b, 1);
            }
        } else if (ch[0] == 'Q') {
            int x1, x2, y1, y2;
            cin >> x1 >> x2 >> y1 >> y2;
            x1++, x2++, y1++, y2++;
            if (x1 > x2)
                swap(x1, x2);
            if (y1 > y2)
                swap(y1, y2);
            cout << sum(x2, y2) - sum(x1 - 1, y2) - sum(x2, y1 - 1) + sum(x1 - 1, y1 -1) << endl;
        } else {
            int a, b;
            cin >> a >> b;
            a++, b++;
            if (flag[a][b]) {
                flag[a][b] = 0;
                update(a, b, -1);
            }
        }
    }
}