[POJ 2155] 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

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

 

题解：

二维树状数组

对于翻转操作：在二维bit中记录翻转的次数，由于是后缀操作，所以要把不在翻转范围内的区间翻回来

对于询问操作：每次查询(x,y)在bit中的前缀和最后%2就是结果

/*
  挑战2-树状数组
  二维bit
  写起来简单，想起来难= =
  by-solution
*/
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;

const int N = 1010;

int n,qu,c[N][N];

int gi() {
  int x=0,o=1; char ch=getchar();
  while(ch!='-' && (ch<'0'||ch>'9')) ch=getchar();
  if(ch=='-') ch=getchar(),o=-1;
  while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
  return o*x;
}

void add(int i, int j) {
  for(int p=i; p<=n; p+=lowbit(p))
    for(int q=j; q<=n; q+=lowbit(q))
      c[p][q]++;
}

int query(int i, int j) {
  int ret=0;
  for(int p=i; p; p-=lowbit(p))
    for(int q=j; q; q-=lowbit(q))
      ret+=c[p][q];
  return ret;
}

int main() {//poj2155
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  int T=gi();
  while(T--) {
    memset(c,0,sizeof(c));
    n=gi(),qu=gi();
    for(int i=1; i<=qu; i++) {
      char ch;
      scanf("%c", &ch);
      if(ch=='C') {
    int x=gi(),y=gi(),xx=gi(),yy=gi();
    add(x,y),add(xx+1,y),add(x,yy+1),add(xx+1,yy+1);
      }
      else {
    int x=gi(),y=gi();
    printf("%d\n", query(x,y)%2);
      }
    }
    printf("\n");
  }
  return 0;
}