HDU 2642 Stars (二维树状数组)

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#二维树状数组

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本文介绍了一个基于二维树状数组的算法，用于解决在一个二维区域内，根据点亮和熄灭星星的操作，快速查询特定子区域内点亮星星数量的问题。通过使用树状数组进行高效更新和查询，该算法能够处理大量操作和查询，适用于天文学爱好者统计天空中星星数量的场景。

Stars

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 2495 Accepted Submission(s): 1034

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

B 581 145

Q 0 600 0 200

D 581 145

Q 0 600 0 200

Sample Output

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2642

题目大意：给一个二维区域，B表示将对应坐标的星星点亮，D表示将对应坐标的星星熄灭，Q询问一个子区域中亮着的星星个数

题目分析：二维树状数组维护即可

#include <cstdio>
#include <algorithm>
using namespace std;
int const MAX = 1005;
int m, x1, y1, x2, y2, sum[MAX][MAX];
bool sta[MAX][MAX];

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

int add(int x, int y, int val) {
    for (int i = x; i < MAX; i += lowbit(i)) {
        for (int j = y; j < MAX; j += lowbit(j)) {
            sum[i][j] += val;
        }
    }
}

int query(int x, int y) {
    int ans = 0;
    for (int i = x; i > 0; i -= lowbit(i)) {
        for (int j = y; j > 0; j -= lowbit(j)) {
            ans += sum[i][j];
        }
    }
    return ans;
}

int main() {
    scanf("%d", &m);
    char s[2];
    while (m--) {
        scanf("%s", s);
        if (s[0] == 'B') {
            scanf("%d %d", &x1, &y1);
            x1++; y1++;
            if (!sta[x1][y1]) {
                add(x1, y1, 1);
                sta[x1][y1] = true;
            }
        } else if (s[0] == 'D') {
            scanf("%d %d", &x1, &y1);
            x1++; y1++;
            if (sta[x1][y1]) {
                add(x1, y1, -1);
                sta[x1][y1] = false;
            }
        } else {
            scanf("%d %d %d %d", &x1, &x2, &y1, &y2);
            x1++; y1++;
            x2++; y2++;
            if (x1 > x2) {
                swap(x1, x2);
            }
            if (y1 > y2) {
                swap(y1, y2);
            }
            printf("%d\n", query(x2, y2) + query(x1 - 1, y1 - 1) - query(x1 - 1, y2) - query(x2, y1 - 1));
        }
    }
}