POJ 2155 Matrix（二维树状数组）

Matrix

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

题目大意：给出一个只由0和1构成的矩阵，初始时矩阵元素全为0。我们可以对矩阵进行取反操作，即将1变为0或将0变为1。现在给出取反操作的左上角元素与右下角元素的坐标，要求对于每次查询操作，输出对应位置元素的值。

解题思路：对于每次取反操作，我们只需在树状数组中更新4个值，分别是(x1,y1),(x2 + 1,y1),(x1,y2 + 1),(x2 + 1,y2 + 1)，查询时只需判断sum(x,y)是否为偶数即可，证明如下

假设插入(x1,y1),(x2 + 1,y1),(x1,y2 + 1),(x2 + 1,y2 + 1)四个值，查询(x,y)，如上图所示。

当(x,y)属于1，2，3，4或7这5个区域时，计算sum(x,y)不受插入的影响；

当(x,y)属于第5个区域时，sum(x,y)会受到(x1,y1)的影响，sum(x,y)会增加1，更改正确；

当(x,y)属于第6个区域时，sum(x,y)会受到(x1,y1)和(x2 + 1,y1)的影响，sum(x,y)会比之前增加2，结果不受影响，更改正确；

当(x,y)属于第8个区域时，sum(x,y)会受到(x1,y1)和(x1,y2 + 1)的影响，sum(x,y)会比之前增加2，结果不受影响，更改正确；

当(x,y)属于第9个区域时，sum(x,y)会受到(x1,y1),(x2 + 1,y1),(x1,y2 + 1)和(x2 + 1,y2 + 1)的影响，sum(x,y)会比之前增加4，结果不受影响，更改正确。

PS:以上图片及证明均来自论文《浅谈信息学竞赛中的“0”和“1”》

代码如下：

#include <algorithm>#include <cctype>#include <climits>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <queue>#include <set>#include <stack>#include <vector>#define EPS 1e-6#define INF INT_MAX / 10#define LL long long#define MOD 100000000#define PI acos(-1.0)using namespace std;const int maxn = 1005;int bit[maxn][maxn];int n;int sum(int posx,int posy){    int s = 0;    int x = posx,y = posy;    while(x > 0){        while(y > 0){            s += bit[x][y];            y -= y & -y;        }        x -= x & -x;        y = posy;    }    return s;}void add(int posx,int posy,int val){    int x = posx,y = posy;    while(x <= n){        while(y <= n){            bit[x][y] += val;            y += y & -y;        }        x += x & -x;        y = posy;    }}int main(){    int t,m,a,b,c,d;    char opt[2];    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        memset(bit,0,sizeof(bit));        while(m--){            scanf("%s",opt);            if(opt[0] == 'C'){                scanf("%d %d %d %d",&a,&b,&c,&d);                add(a,b,1);                add(c + 1,b,1);                add(a,d + 1,1);                add(c + 1,d + 1,1);            }            if(opt[0] == 'Q'){                scanf("%d %d",&a,&b);                printf("%d\n",sum(a,b) % 2 ? 1 : 0);            }        }        if(t)            printf("\n");    }    return 0;}