Poj2155 Matrix 树状数组

题目描述：

Matrix Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 26468 Accepted: 9778

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

Source

POJ Monthly,Lou Tiancheng

思路：

因为是改变一个矩阵求点，可以联想到树状数组的该段求点的算法。

只用加一层循环，相当于横着竖着都是一个一维树状数组。

关于add函数：

改变矩阵（x1,x2,y1,y2）所需的操作就是add(x1,y1,1),add(x1,y2+1,-1),add(x2+1,y1,-1),add(x2+1,y2+1,1)。

代码：

#include<cstdio> #include<iostream> #include<cstring> using namespace std; int n,m; int a[1010][1010]= {0}; int lowbit(int x) { return x&(-x); } void add(int x,int y,int z) { for(int i=x; i<=n; i+=lowbit(i)) for(int j=y; j<=n; j+=lowbit(j)) a[i][j]+=z; } int getsum(int x,int y) { int s=0; for(int i=x; i>0; i-=lowbit(i)) { for(int j=y; j>0; j-=lowbit(j)) { s+=a[i][j]; } } return s; } int main() { int t; scanf("%d",&t); while(t--) { memset(a,0,sizeof(a)); scanf("%d%d",&n,&m); for(int i=1; i<=m; i++) { char ask; scanf("%*c%c",&ask); if(ask=='C') { int x1,y1,x2,y2; scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(x1,y1,1),add(x1,y2+1,-1),add(x2+1,y1,-1),add(x2+1,y2+1,1); } if(ask=='Q') { int x,y; scanf("%d%d",&x,&y); printf("%d\n",getsum(x,y)%2); } } printf("\n"); } return 0; }