poj 2155 Martrix（二维树状数组）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10282		Accepted: 3859

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

  

          此题大意：给你一个2维平面，C:(X1,Y1)(X2,Y2)就是将这里范围的点改变：0->1、1->0，Q:(x,y)就是问这个点是什么。

      典型的二维树状数组，就是增加一维，写法和一维基本一致，不多说，看下面的吧！！！

链接：http://poj.org/problem?id=2155

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

int a[1005][1005], n;

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

void update(int i, int j, int x)    //二维树状数组更新
{
    int tj;
    while(i <= n)
    {
        tj = j;
        while(tj <= n)
        {
            a[i][tj] += x;
            tj += lowbit(tj);
        }
        i += lowbit(i);
    }
}

int sum(int i, int j)   //二维树状数组求和
{
    int tj, sum = 0;
    while(i > 0)
    {
        tj = j;
        while(tj > 0)
        {
            sum += a[i][tj];
            tj -= lowbit(tj);
        }
        i -= lowbit(i);
    }
    return sum;
}

int main()
{
    int T, i, j, t, x1, y1, x2, y2;
    char ch[2];
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &n, &t);
        for(i = 1; i <= n; i++)
            for(j = 1; j <= n; j++)
                a[i][j] = 0;
        while(t--)
        {
            scanf("%s", ch);
            if(ch[0] == 'C')
            {
                scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
                //更新区域-容斥原理
                update(x2+1, y2+1, 1);
                update(x2+1, y1, 1);
                update(x1, y2+1, 1);
                update(x1, y1, 1);
            }
            if(ch[0] == 'Q')
            {
                scanf("%d %d", &x1, &y1);
                printf("%d\n", sum(x1, y1)&1);  //该点的值就是sum(x,y)
            }
        }
        printf("\n");
    }

    return 0;
}