poj 2155 2 차원 트 리 배열 / 구간 업데이트 단일 조회

Matrix
Time Limit: 3000MS
 
Memory Limit: 65536K
Total Submissions: 20002
 
Accepted: 7481
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
직사각형 구간 과 2 차원 트 리 배열 템 플 릿 문제.선분 트 리 로 선분 트 리 를 끼 워 넣 어도 되 고 직사각형 트 리 (상수 가 너무 크다) 는 tle 입 니 다.

#include 
#include 
#include 
#include 
#include 

using namespace std;

#define maxn 1001

int arr[maxn][maxn];
int n,t;

int lowbit(int i)
{
    return i&-i;
}

void add(int x,int y,int v) {
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
            arr[i][j]+=v;
}
int sum(int x,int y){ //    
    int res=0;
    for(int i=x;i;i-=lowbit(i))
        for(int j=y;j;j-=lowbit(j))
            res+=arr[i][j];
    return res;
}

void update(int l,int d,int r,int u,int v){     //    
    add(l,d,v);add(l,u+1,-v);add(r+1,d,v);add(r+1,u+1,-v);
}
int main()
{
    int x;
    scanf("%d", &x);
    while(x--){
        memset(arr,0,sizeof(arr));
        scanf("%d%d",&n,&t);
        char op[3];
        while(t--){
            scanf("%s", op);
            int x1, x2,y1,y2;
            scanf("%d%d", &x1, &y1);
            if(op[0] == 'C'){
                scanf("%d%d", &x2, &y2);
                update(x1,y1,x2,y2,1);
            }
            else{
                printf("%d
", sum(x1,y1)&1);
            }
        }
        printf("
");
    }
    return 0;
}