POJ2155 Matrix <树套树/二维树状数组>

Problem

Matrix

Time Limit: $3000MS$
Memory Limit: $65536K$

Description

Given an $N\times N$ 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 $A[i, j] = 0 (1 \le i, j\le< N)$.
We can change the matrix in the following way. Given a rectangle whose upper-left corner is $(x_1, y_1)$ and lower-right corner is $(x_2, y_2)$, 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. $C$ $x_1$ $y_1$ $x_2$ $y_2$ ($1\le x_1\le x_2\le n, 1\le y_1\le y_2\le n$) changes the matrix by using the rectangle whose upper-left corner is $(x_1, y_1)$ and lower-right corner is $(x_2, y_2)$.
2. $Q$ $x$ $y$ ($1\le x, y\le n$) querys $A[x, y]$.

Input

The first line of the input is an integer $X (X \le 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 $T$ ($2 \le N \le 1000$, $1 \le T \le 50000$) 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 “$C$ $x_1$ $y_1$ $x_2$ $y_2$”, 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

标签：线段树套线段树/二维树状数组

Solution

此题最简单的方法是二维树状数组，但因为二维树状数组没太大用，所以练习线段树的树套树。
此题用作树套树的模板题再合适不过。

Code

#include <iostream>
#include <cstdio>
#include <cstring>
#define MAX_N 1000
using namespace std;
int n, m;
int tr[(MAX_N<<2)+5][(MAX_N<<2)+5];
void modify_y(int v1, int v2, int s, int t, int l, int r) {
    if (s >= l && t <= r) {
        tr[v1][v2] ^= 1;
        return;
    }
    int mid = s+t>>1;
    if (l <= mid)	modify_y(v1, v2<<1, s, mid, l, r);
    if (r >= mid+1)	modify_y(v1, v2<<1|1, mid+1, t, l, r);
}
void modify_x(int v, int s, int t, int x1, int y1, int x2, int y2) {
    if (s >= x1 && t <= x2) {
        modify_y(v, 1, 1, n, y1, y2);
        return;
    }
    int mid = s+t>>1;
    if (x1 <= mid)	modify_x(v<<1, s, mid, x1, y1, x2, y2);
    if (x2 >= mid+1)	modify_x(v<<1|1, mid+1, t, x1, y1, x2, y2);
}
int query_y(int v1, int v2, int s, int t, int pos) {
    if (s == t)	return tr[v1][v2];
    int mid = s+t>>1;
    if (pos <= mid)	return tr[v1][v2]^query_y(v1, v2<<1, s, mid, pos);
    else	return tr[v1][v2]^query_y(v1, v2<<1|1, mid+1, t, pos);
}
int query_x(int v, int s, int t, int x, int y) {
    if (s == t)	return query_y(v, 1, 1, n, y);
    int mid = s+t>>1;
    if (x <= mid)	return query_y(v, 1, 1, n, y)^query_x(v<<1, s, mid, x, y);
    else	return query_y(v, 1, 1, n, y)^query_x(v<<1|1, mid+1, t, x, y);
}
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        memset(tr, 0, sizeof(tr));
        scanf("%d%d", &n, &m);
        while (m--) {
            char ch;
            cin >> ch;
            if (ch == 'C') {
                int x1, y1, x2, y2;
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                modify_x(1, 1, n, x1, y1, x2, y2);
            }
            if (ch == 'Q') {
                int x, y;
                scanf("%d%d", &x, &y);
                printf("%d\n", query_x(1, 1, n, x, y));
            }
        }
        printf("\n");
    }
    return 0;
}