POJ 2155 Matrix

最新推荐文章于 2022-12-28 11:12:37 发布

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本文介绍了一种使用二维树状数组解决特定矩阵操作问题的方法。针对一个N*N的矩阵，通过“not”操作实现0和1的反转，并支持两种指令：一是矩形区域内的元素反转，二是查询指定位置的元素值。文章提供了完整的C++代码实现。

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].

【题目分析】
还是一道区间查询的题目，看到之后很容易想到用二维的树状数组去求解，但是这道题询问的是关于1、0这两种情况，而且还要求01反转。很容易，%2求解就可以了。

【代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1010;
int lowbit(int x){return x&(-x);}
int c[MAXN][MAXN];
int n;
int sum(int x,int y)
{
    int ret = 0;
    for(int i = x;i > 0;i -= lowbit(i))
        for(int j = y;j > 0;j -= lowbit(j))
            ret += c[i][j];
    return ret;
}
void add(int x,int y,int val)
{
    for(int i = x;i <= n;i += lowbit(i))
        for(int j = y;j <= n;j += lowbit(j))
            c[i][j] += val;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int q;
        scanf("%d%d",&n,&q);
        memset(c,0,sizeof(c));
        char op[10];
        int x1,y1,x2,y2;
        while(q--)
        {
            scanf("%s",op);
            if(op[0] == 'C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,1);
                add(x2+1,y1,1);
                add(x1,y2+1,1);
                add(x2+1,y2+1,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                if(sum(x1,y1)%2 == 0)printf("0\n");
                else printf("1\n");
            }
        }
        if(T > 0)printf("\n");
    }
    return 0;
}