poj 2155 Matrix(二维树状数组)

最新推荐文章于 2025-01-03 19:49:02 发布

Timeclimber

最新推荐文章于 2025-01-03 19:49:02 发布

阅读量245

点赞数

CC 4.0 BY-SA版权

分类专栏：树状数组文章标签： hdu 二维树状数组

本文链接：https://blog.youkuaiyun.com/Timeclimber/article/details/79062989

树状数组专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一种解决二维矩阵查询与更新问题的有效方法。通过使用二维BIT（Binary Indexed Tree）来快速实现矩阵区域的反转操作及单点查询，提供了一个简洁且高效的C++实现方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 30316		Accepted: 11094

Description

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N).

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case.

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above.

Output

For each querying output one line, which has an integer representing A[x, y].

There is a blank line between every two continuous test cases.

Sample Input

Sample Output

Source

POJ Monthly,Lou Tiancheng

题意：

给了一个长宽为N的矩阵，C操作可以把以(x1,y1),(x2,y2)为对角线的矩阵内的每一个点进行反转操作。Q是查询(x,y)点的值。

思路：

参考https://wenku.baidu.com/view/1e51750abb68a98271fefaa8。很巧妙的方法。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int sum[1005][1005];
int t,n,m,x1,x2,y1,y2;
int lowbit(int i)
{
    return i&(-i);
}
void update(int x,int y){
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=n;j+=lowbit(j))
        {
            sum[i][j]++;
        }
    }
}
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%2;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        char s[5];
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",s);
            if(s[0]=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                x1++,x2++,y1++,y2++;
                update(x2,y2);
                update(x2,y1-1);
                update(x1-1,y2);
                update(x1-1,y1-1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                printf("%d\n",query(x1,y1));
            }
        }
        printf("\n");
    }
    return 0;
}