Matrix(poj2155)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25139		Accepted: 9314

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

思路：二维树状数组；

http://download.csdn.net/detail/lenleaves/4548401

这个解释的很好；

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<queue>
#include<string.h>
using namespace std;
int bit[1005][1005];
int lowbit(int x)
{
        return x&(-x);
}
void add(int x,int y)
{
        int i,j;
        for(i = x; i <= 1000; i+=lowbit(i))
        {
                for(j = y; j <= 1000; j+=lowbit(j))
                {
                        bit[i][j]+=1;
                        bit[i][j]%=2;
                }
        }
}
int ask(int x,int y)
{
        int i,j;
        int sum = 0;
        for(i = x; i > 0; i-=lowbit(i))
        {
                for(j = y; j > 0; j-=lowbit(j))
                {
                        sum += bit[i][j];
                }
        }
        return sum%2;
}
int main(void)
{
        int T;
        scanf("%d ",&T);
        while(T--)
        {
                memset(bit,0,sizeof(bit));
                int i,j;
                int N,q;
                scanf("%d %d ",&N,&q);
                char a[10];
                while(q--)
                {
                        scanf("%s",a);
                        int x,y,x1,y1;
                        if(a[0] == 'C')
                        {
                                scanf("%d %d %d %d",&x,&y,&x1,&y1);
                                add(x,y);
                                add(x1+1,y1+1);
                                add(x,y1+1);
                                add(x1+1,y);
                        }
                        else
                        {
                                scanf("%d %d",&x,&y);
                                int ac = ask(x,y);
                                printf("%d\n",ac);
                        }
                }
                printf("\n");
        }
        return 0;
}