POJ-2155-Matrix (2-dimensional tree array)

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

Idea: I think this idea is too powerful when I see other people's code. I use a two-dimensional tree array to maintain the number of vertex occurrences. Every time I add a rectangle, add the four vertices (pay attention to the boundary issue, x2, y2 must be added ). When querying, calculate the total number of vertices from () to (x, y). The odd number is 1, and the even number is 0.

#include 
 
  int sum[1005][1005],n,m;int lowbit(int x){    return x&-x;}void update(int x,int y){    for(int i=x;i<=n;i+=lowbit(i))    {        for(int j=y;j<=n;j+=lowbit(j))        {            sum[i][j]++;        }    }}int query(int x,int y){    int t=0;    for(int i=x;i>0;i-=lowbit(i))    {        for(int j=y;j>0;j-=lowbit(j))        {            t+=sum[i][j];        }    }    return t&1;}int main(){    int T,i,j,x1,y1,x2,y2;    char s[5];    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++) for(j=1;j<=n;j++) sum[i][j]=0;        while(m--)        {            scanf("%s",s);            if(s[0]=='C')            {                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);                update(x1,y1);                update(x1,y2+1);                update(x2+1,y2+1);                update(x2+1,y1);            }            else            {                scanf("%d%d",&x1,&y1);                printf("%d\n",query(x1,y1));            }        }        printf("\n");    }}