题目链接：http://poj.org/problem?id=2155
Matrix
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 32950 | Accepted: 11943 |
Description
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 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
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
题意概括：
有一个初始值为0的N*N的二维矩阵，有T次操作，每次操作有两种选择：
C : 修改以（x1, y1）为左上角（x2, y2）为右下角的矩阵的值（0和1互换）
Q：查询（x, y）的值为 0 或者 为 1；
解题思路：
涉及到多次区间修改和区间查询的优先考虑线段树和树状数组，这道题巧妙之处在于灵活运用树状数组的前缀和，把区间修改转换单点修改，一维需要标记两个点而二维需要标记四个点，一个点用于发挥效果，另外三个点用于消除效果，因为树状数组维护的是前缀和，而对于二维树状数组，查询的则是以（1，1）为左上角，（x，y）为右下角的矩阵的和。
第二就是我们可以借助修改次数的奇偶性来判断该点的值为 0 / 1；
例如：N = 3；C：x1 = 1, y1 = 1, x2 = 2, y2 = 2;
| 1 | 1 | |
| （x，y） | ||
| 1 | 1 |
(因为我们只需要知道奇偶性，所以矩阵外的点加1即可消除效果）
AC code：
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define ll long long int;
#define INF 0x3f3f3f3f
using namespace std;
const int MAXN = 1e3+; int mmp[MAXN][MAXN];
int N, T; int lowbit(int x)
{
return x&(-x);
} void add(int x, int y, int value)
{
for(int i = x; i <= N; i += lowbit(i))
for(int j = y; j <= N; j += lowbit(j))
mmp[i][j]+=value;
} int sum(int x, int y)
{
int res = ;
for(int i = x; i > ; i -= lowbit(i))
for(int j = y; j > ; j -= lowbit(j))
res+=mmp[i][j];
return res;
} void init()
{
for(int i = ; i <= N; i++)
for(int j = ; j <= N; j++)
mmp[i][j] = ;
} int main()
{
int T_case;
char com[];
int x, y, x1, x2, y1, y2;
scanf("%d", &T_case);
while(T_case--)
{
scanf("%d%d", &N, &T);
init();
while(T--)
{
scanf("%s", &com);
if(com[] == 'C')
{
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
add(x1, y1, );
add(x2+, y1, );
add(x1, y2+, );
add(x2+, y2+, );
}
else if(com[] == 'Q')
{
scanf("%d%d", &x, &y);
int res = ;
res = sum(x, y);
// printf("res: %d\n", res);
if(res%) printf("1\n");
else printf("0\n");
}
}
puts("");
}
return ;
}