poj2155一个二维树状数组

                                                                                                               Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15125		Accepted: 5683

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

一个二维树状数组，好像题目说错了，其实是左下角和右上角的。

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;
long long int a[1100][1100];
int m;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int x1,int y1)
{
    int i,j;
    for(i=x1;i>0;i-=lowbit(i))
    for(j=y1;j>0;j-=lowbit(j))
    a[i][j]++;

    for(i=x-1;i>0;i-=lowbit(i))
    for(j=y-1;j>0;j-=lowbit(j))
    a[i][j]++;

    for(i=x1;i>0;i-=lowbit(i))
    for(j=y-1;j>0;j-=lowbit(j))
    a[i][j]++;

    for(i=x-1;i>0;i-=lowbit(i))
    for(j=y1;j>0;j-=lowbit(j))
    a[i][j]++;
}
int fun(int x,int y)
{
    int i,j;
    int sum=0;
    for(i=x;i<=m;i+=lowbit(i))
    for(j=y;j<=m;j+=lowbit(j))
    sum+=a[i][j];
    return sum%2;
}
int main()
{
    //freopen("int.txt","r",stdin);
    int n;
    int i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        memset(a,0,sizeof(a));
        int k;
        scanf("%d%d",&m,&k);
        char b;
        for(j=0;j<k;j++)
        {
            cin>>b;
            int x1,x2,y1,y2;
            if(b=='Q')
            {
                scanf("%d%d",&x1,&y1);
                printf("%d
",fun(x1,y1));
            }
            else
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                update(x1,y1,x2,y2);
            }
        }
        if(i!=n-1)
        printf("
");
    }
}