poj 2155 二维树状数组 二维线段树

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

二维树状数组。两个for循环就好

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <vector>
#define lowbit(x) ((x)&(-x))
using namespace std;
const int maxn=1010;
int c[maxn][maxn];
int n;
void update(int x,int y)
{
    int i,k;
    for(i=x;i<=n;i+=lowbit(i))
    {
        for(k=y;k<=n;k+=lowbit(k))
            c[i][k]++;
    }
}
int Get(int x,int y)
{
       int i,k,sum=0;
       for(i=x;i>0;i-=lowbit(i))
       {
        for(k=y;k>0;k-=lowbit(k))
            sum+=c[i][k];
       }    
       return sum;
}
int main()
{
    int t;
    int x1,x2,y1,y2;
     scanf("%d",&t);
        int m;
        while(t--)
        {
            memset(c,0,sizeof(c));
            scanf("%d%d",&n,&m);
            while(m--)
            {
                char ch;
                scanf(" %c",&ch);
                if(ch=='C') 
                {
                    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                    x1++,x2++,y1++,y2++;  //因为更新的时候下面的坐标有的减一所以得从二开始
                    update(x2,y2);
                    update(x1-1,y1-1); 
                    update(x1-1,y2);
                    update(x2,y1-1);

                }
                else
                {
                    scanf("%d%d",&x1,&y1);
                    printf("%d\n",Get(x1,y1)%2);
                }
            }
            printf("\n");
        }

}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define xlson kx<<1,xl,mid
#define xrson kx<<1|1,mid+1,xr
#define ylson ky<<1,yl,mid
#define yrson ky<<1|1,mid+1,yr
#define MAXN 1005
#define mem(a) memset(a,0,sizeof(a));
bool tree[MAXN<<2][MAXN<<2];
int X,N,T;
int num,X1,X2,Y1,Y2;
char ch;
void editY(int kx,int ky,int yl,int yr)
{
    if(Y1<=yl&&yr<=Y2)
    {
        tree[kx][ky]=!tree[kx][ky];
        return ;
    }
    int mid=(yl+yr)>>1;
    if(Y1<=mid) editY(kx,ylson);
    if(Y2>mid) editY(kx,yrson);
}

void editX(int kx,int xl,int xr)
{
    if(X1<=xl&&xr<=X2)
    {
        editY(kx,1,1,N);
        return ;
    }
    int mid=(xl+xr)>>1;
    if(X1<=mid) editX(xlson);
    if(X2>mid) editX(xrson);
}

void queryY(int kx,int ky,int yl,int yr)
{
    if(tree[kx][ky]) num++;//总的全部面积有没有被翻过，就是父区间是否被翻过，先统计父区间
                      //再统计子区间。
    if(yl==yr) return ;
    int mid=(yl+yr)>>1;
    if(Y1<=mid) queryY(kx,ylson);
    else queryY(kx,yrson);
}

void queryX(int kx,int xl,int xr){
        queryY(kx,1,1,N);
        if(xl==xr) return ;
        int mid=(xl+xr)>>1;
        if(X1<=mid) queryX(xlson);
        else queryX(xrson);
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    while(t--)
    {
        scanf("%d%d",&N,&T);
        mem(tree);
        for(int i=0;i<T;i++)
        {
            scanf(" %c%d%d",&ch,&X1,&Y1);
            if(ch=='C') 
            {
                scanf("%d%d",&X2,&Y2);
                editX(1,1,N);
            }
            else
            {
                num=0;
                queryX(1,1,N);
                if(num&1) printf("1\n");
                else printf("0\n");
            }
        }
        if(t) printf("\n"); 
    }

}