Matrix

【题目描述】
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].
【输入格式】
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.
【输出格式】
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.
【样例输入】

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

【样例输出】

1001

【分析】
二维差分+树状数组

#include<cstdio>#include<cstring>using namespace std;const int Max=1010;int bit[Max][Max]={0},n;int lowbit(int x){return x&(-x);}void Add(int x,int y,int z){    for(int i=x;i>0;i-=lowbit(i))        for(int j=y;j>0;j-=lowbit(j))            bit[i][j]=(bit[i][j]+z+2&1);}int Query(int x,int y){    int sum=0;    for(int i=x;i<=n;i+=lowbit(i))        for(int j=y;j<=n;j+=lowbit(j))            sum+=bit[i][j];    return sum&1;}int main(){    int t,q,xx1,xx2,yy1,yy2;    char str[10];    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&q);        memset(bit,0,sizeof(bit));        while(q--){            scanf("%s",str);            if(str[0]=='C'){                scanf("%d %d %d %d",&xx1,&yy1,&xx2,&yy2);                Add(xx2,yy2,1),Add(xx1-1,yy2,-1),Add(xx2,yy1-1,-1),Add(xx1-1,yy1-1,1);            }            else{                scanf("%d %d",&xx1,&yy1);                printf("%d\n",Query(xx1,yy1));            }        }        printf("\n");    }}