poj2155二维树状数组

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

题意：给一个0，1矩阵，有两种操作一种是把子矩阵进行非，一种是求子矩阵是否有1

题解：二维树状数组，用

add(x1,y1);
add(x1,y2+1);
add(x2+1,y1);
add(x2+1,y2+1);
求解时%2即可

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-9;
const int N=1000+5,maxn=32000+5,inf=0x3f3f3f3f;

int c[N][N];

void add(int x1,int y1)
{
    for(int i=x1;i<N;i+=i&(-i))
        for(int j=y1;j<N;j+=j&(-j))
           c[i][j]++;
}
int sum(int x,int y)
{
    int ans=0;
    for(int i=x;i>0;i-=i&(-i))
        for(int j=y;j>0;j-=j&(-j))
           ans+=c[i][j];
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int n;
    int t,k;
    cin>>t;
    while(t--){
        cin>>n>>k;
        memset(c,0,sizeof c);
        while(k--){
            string s;
            cin>>s;
            if(s[0]=='C')
            {
                int x1,x2,y1,y2;
                cin>>x1>>y1>>x2>>y2;
                add(x1,y1);
                add(x1,y2+1);
                add(x2+1,y1);
                add(x2+1,y2+1);
            }
            else
            {
                int x,y;
                cin>>x>>y;
                cout<<sum(x,y)%2<<endl;
            }
        }
        cout<<endl;
    }
    return 0;
}