Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 20599 Accepted: 7673
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
二维树状数组的经典例题
二维数组
题意：一个n*n的矩阵，矩阵里的每一个数要么是0，要么是1。
有两种操作，C操作是将区间[(x1,y1),(x2,y2)]中所有的数反转（0变1,1变0），Q操作是查询f[x][y]的数值。
分析：
根据这个题目中介绍的这个矩阵中的数的特点不是1就是 0，这样我们只需记录每个格子改变过几次，即可判断这个格子的数字。
怎么记录改变过几次呢？
我们先看一维的
0 0 0 0 0
【1，3】
1 0 0 -1 0
则i点相当于改变sum(i)次
上升到二维也是一样的，在纸上模拟一下就ok。
所以就变成了区间修改+查询前缀和的问题的问题
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define N 1040
#define ll long long
using namespace std;
int n;
int bit[N][N];
int sum(int i,int j) ///查询单点的值
{
int s=0;
while(i>0) {
int jj=j;
while(jj>0) {
s+=bit[i][jj];
jj-=jj&-jj;
}
i-=i&-i;
}
return s;
}
void add(int i,int j,int x) {
while(i<=n) {
int jj=j;
while(jj<=n) {
bit[i][jj]+=x;
jj+=jj&-jj;
}
i+=i&-i;
}
}
int main()
{
int t;
cin>>t;
while(t--) {
int q;
scanf("%d%d ",&n,&q);
memset(bit,0,sizeof bit);
char c;
int x,y,x1,y1;
while(q--) {
scanf("%c",&c);
if(c=='C') {
scanf("%d%d%d%d",&x,&y,&x1,&y1);
add(x,y,1);
add(x,y1+1,-1);
add(x1+1,y,-1);
add(x1+1,y1+1,1);//重叠的部分加上
} else {
scanf("%d%d",&x,&y);
cout<<sum(x,y)<<endl;
printf("%d\n",sum(x,y)%2);
}
getchar();
}
if(t)printf("\n");
}
return 0;
}
