POJ 2155 Matrix

二维树状数组。。。。

                         Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 15575		Accepted: 5854

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

Source

POJ Monthly,Lou Tiancheng 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int tree[2000][2000],n,m,t;

inline int lowbit(int x)
{
    return -x&x;
}

void update(int x,int y,int v)
{
    int temp=y;
    while(x<=n)
    {
        y=temp;
        while(y<=n)
        {
            tree[x][y]+=v;
            y+=lowbit(y);
        }
        x+=lowbit(x);
    }
}

int getsum(int x,int y)
{
    int sum=0;
    int temp=y;
    while(x>0)
    {
        y=temp;
        while(y>0)
        {
            sum+=tree[x][y];
            y-=lowbit(y);
        }
        x-=lowbit(x);
    }
    return sum;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(tree,0,sizeof(tree));
        scanf("%d%d",&n,&m);
        n++;char str[3];
        while(m--)
        {
            scanf("%s",str);
            if(str[0]=='C')
            {
                int x1,x2,y1,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                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 if(str[0]=='Q')
            {
                int x,y;
                scanf("%d%d",&x,&y);
                x++;y++;
                printf("%d
",getsum(x,y)&1);
            }
        }
        putchar(10);
    }
    return 0;
}