POJ 2155 Matrix（二维树状数组）

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

大致题意：给你一个n*n的矩阵，初始值都为0，然后又若干次操作，第一种，给你左上角坐标和右下角坐标，将这个子矩形中的值翻转0变成1，1变成0，第二种，给你一个坐标让你求出该位置此时的值是多少。

思路：想要知道某个位置此时的值是0还是1，我们只需记录该位置被翻转了多少次即可，先考虑一维的情况，假设我们需要将i到j之间的位置翻转一遍，我们只需在i号位置上add（i，1），j+1号位置上add（j+1，-1）即可，然后查询x位置上的值时用树状数组求下1到x区间的和即可，那么对于二维，同理，假设我们需要将（x1，y1）和（x2，y2）这个区间内的值翻转，我们只需add（x1，y1，1）
add（x1，y2+1，-1）
add（x2+1，y1，-1）
add（x2+1，y2+1，1）
如此即可。询问（x，y）处的值时，只需对（1，1）到（x，y）这个子区间求下和即可。

代码如下

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std; const int N=1e3+5;int C[N][N];int n;int lowbit(int x){    return x&-x;}void add(int x,int y,int date){    for(int i=x;i<=n;i+=lowbit(i))    for(int j=y;j<=n;j+=lowbit(j))        C[i][j]+=date;}int sum(int x,int y){    int ans=0;    for(int i=x;i>0;i-=lowbit(i))    for(int j=y;j>0;j-=lowbit(j))        ans+=C[i][j];    return ans;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(C,0,sizeof(C));        int q;        scanf("%d%d",&n,&q);        char ch[10];        int x,y,xx,yy;        while(q--)        {            scanf("%s",ch);            if(ch[0]=='C')            {                scanf("%d%d%d%d",&x,&y,&xx,&yy);                add(x,y,1);                add(x,yy+1,-1);                add(xx+1,y,-1);                add(xx+1,yy+1,1);            }            else             {                scanf("%d%d",&x,&y);                printf("%d\n",sum(x,y)%2);            }        }        if(T>0) printf("\n");    }    return 0; }