POJ 2155 Matrix 树状数组

Matrix

Problem 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. <br> <br>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. <br>

 

Output

For each querying output one line, which has an integer representing A[x, y]. <br> <br>There is a blank line between every two continuous test cases. <br>

 

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

 

//题意：有个N*N的矩阵，每个点的值要么是0，要么是1，每个点一开始的初始值为0。现在有2种操作：“C”：要输入x1,y1,x2,y2(x2>x1 , y2>y1)，把以（x1,y1），（x2,y2）为对角线顶点的那个矩形里的值改变（若原来是1，变为0；原来是0，变为1）。“Q”：输入x,y，要求这点的值是0还是1。

//思路：最先想到的应该是定义一个map二维数组，在根据要求去改变map里的值，但只要稍微想一下就知道这么暴力（N*N）绝逼TLE，于是就考虑怎样高效地改变一个区间里的值，那么显然树状数组（复杂度N*logN）是一个好方法。

然后这里还有一个小技巧，每次1->0,0->1很麻烦，其实只要每个点要改变的时候都+1就行了，最后要计算的时候，如果是奇数就是1，偶数就是0。

还有“C”的时候，x2,y2都要+1（即（x2+1,y2+1））,这是考虑到顶点的情况，自己画个图感受下就很清楚了。

（搞清楚树状数组推荐博客：http://blog.csdn.net/int64ago/article/details/7429868）

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <cstdlib>#include <algorithm>using namespace std;const int MAX = 1010;int n, num;int c[MAX][MAX];int lowbit(int x){return x&(-x);}void add(int x,int y, int val){for (int i = x; i <= n; i += lowbit(i)){for (int j = y; j <= n; j += lowbit(j)){c[i][j] += val;}}}int sum(int x, int y){int sum = 0;for (int i = x; i > 0; i -= lowbit(i)){for (int j = y; j > 0; j -= lowbit(j)){sum += c[i][j];}}return sum;}int main(){int i, j, k;int T;cin >> T;while (T--){char a;int x1, y1, x2, y2;scanf("%d%d", &n, &num);getchar();memset(c, 0, sizeof(c));for (i = 1; i <= num; i++){scanf("%c", &a);if (a == 'C'){scanf("%d%d%d%d", &x1, &y1, &x2, &y2);//用树状数组只要传入4个角，遍历的复杂度是NlogN//暴搜的话是N平方，肯定TLE//自己画个图更好理解一点//x2,y2都+1是为了考虑4个顶点的情况//改变的每个坐标的值不用真的从0->1,1->0//每次都+1就好了，如果是偶数，表明这点就是0；奇数的话就是1；add(x2 + 1, y2 + 1, 1);add(x1, y1, 1);add(x1, y2 + 1, 1);add(x2 + 1, y1, 1);}else if (a == 'Q'){scanf("%d%d", &x1, &y1);int temp = sum(x1, y1) % 2;printf("%d\n", temp);}getchar();}if (T != 0)printf("\n");}return 0;}