【poj2155】Matrix(二维树状数组区间更新+单点查询)

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

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1

 

【题意】

n*n坐标图起初都为0，C：翻转左下和右上两个坐标围成的矩阵中所有点，Q：查询此点的0 1状态

【思路】

二维树状数组区间修改+单点查询模板题，先考虑一维。

一维单点查询就是前缀和，即query(x)。

区间修改先让s-n都加num，再让t+1-n减去num，即update(s, num)，update(t+1, -num)。

二维的单点查询变成二维就好了query(x, y)。

区间修改update(x1, y1, num), update(x2+1, y1, -num), update(x1, y2+1, -num), update(x2+1, y2+1, num)。开始也是黑人问号，在纸上画几笔就看出来了。

这道题问得是0 1状态，那么只要统计翻转次数是偶次还是奇次就行了

【代码】

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