POJ 2155 Matrix 二维线段树

给出一个只包含0和1的矩阵，要求支持子矩阵整体异或1和查询单点。

发现异或特性，所以省的另外开个lazy标记了。直接修改seg就可以了。所谓的标记永久化？

回调函数还不错233，省的复制代码看起来很恶心（应该是叫回调函数吧，逃）。

#include <cstdio>
#include <cstring>
bool data[2300][2300];

struct SegTree {
void (*modify_callback)(int i, int l, int r, int a, int b);
void (*query_callback)(int i, int l, int r, int a);
void modify(int i, int l, int r, int ql, int qr) {
if (l == ql && r == qr) {
modify_callback(i, l, r, ql, qr);
return;
}
int mid = l + r >> 1;
if (qr <= mid) modify(i * 2, l, mid, ql, qr);
else if (ql > mid) modify(i * 2 + 1, mid + 1, r, ql, qr);
else modify(i * 2, l, mid, ql, mid), modify(i * 2 + 1, mid + 1, r, mid + 1, qr);
}
void query(int i, int l, int r, int a) {
query_callback(i, l, r, a);
if (l == r) return;
int mid = l + r >> 1;
if (a <= mid) query(i * 2, l, mid, a);
else query(i * 2 + 1, mid + 1, r, a);
}
} segX, segY;
int x1, x2, y1, y2, operation_last, ans = 0, n;
void modify_Y_callback(int i, int l, int r, int a, int b) {
data[operation_last][i] ^= 1;
}
void modify_X_callback(int i, int l, int r, int a, int b) {
operation_last = i;
segY.modify(1, 1, n, y1, y2);
}
void query_Y_callback(int i, int l, int r, int a) {
ans ^= data[operation_last][i];
}
void query_X_callback(int i, int l, int r, int a) {
operation_last = i;
segY.query(1, 1, n, y1);
}
int main() {
segX.modify_callback = modify_X_callback; segX.query_callback = query_X_callback;
segY.modify_callback = modify_Y_callback; segY.query_callback = query_Y_callback;
int t, m; char op[2];
scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
memset(data, 0, sizeof data);
while (m--) {
scanf("%s", op);
if(*op == 'C') {
scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
segX.modify(1, 1, n, x1, x2);
} else {
ans = 0;
scanf("%d%d", &x1, &y1);
segX.query(1, 1, n, x1);
printf("%d\n", ans);
}
}
putchar('\n');
}
return 0;
}

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 22291		Accepted: 8294

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