Matrix（二维数状数组+求和+区间更新+单点查询）

Problem P

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 65   Accepted Submission(s) : 27

Problem 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. <br> <br>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. <br>

 

Output

For each querying output one line, which has an integer representing A[x, y]. <br> <br>There is a blank line between every two continuous test cases. <br>

 

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：

给出矩阵左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1，然后给出询问，问某个点是多少？

思路：

此题的关键就是要知道单点查询就是求和了。因为初始化全为0，题目告诉我们不是1就是0这个特点，这样我们只需记录每个格子改变过几次（可以%2），即可判断这个格子的数字。注意格式的输出（QAQ）

代码：

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
#define maxn 1010
using namespace std;
int a[maxn][maxn];
int N;
int lowbit(int i)
{
return i&-i;
}
int sum(int i,int j) //求和
{
int summ=0;
while(i>0){
int jj=j;
while(jj>0){
summ+=a[i][jj];
jj-=lowbit(jj);
}
i-=lowbit(i);
}
return summ;
}
void add(int i,int j,int x){ //区间更新
while(i<=N){
int jj=j;
while(jj<=N){
a[i][jj]+=x;
jj+=lowbit(jj);
}
i+=lowbit(i);
}
}
int main()
{
int X,T;
int i,j;
int x1,x2,y1,y2;
char ch;
scanf("%d",&X);
while(X--)
{
memset(a,0,sizeof(a));
scanf("%d%d",&N,&T);
for(i=1;i<=T;i++)
{
scanf("%s",&ch);
if(ch=='C')
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
add(x1,y1,1);
add(x1,y2+1,-1);
add(x2+1,y1,-1);
add(x2+1,y2+1,1); //加上减去两次的部分
}
else
{
scanf("%d%d",&x1,&y1);
printf("%d\n",sum(x1,y1)%2);
}

}
printf("\n");
}
return 0;
}