POJ2155 Matrix <树套树/二维树状数组>

Matrix
Time Limit: 3000MS Memory Limit: 65536K

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

标签：线段树套线段树/二维树状数组

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。

AC代码如下：

#include <iostream>#include <cstdio>#include <cstring>#define MAX_N 1000using 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;}