POJ2155 树状数组

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 26650		Accepted: 9825

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

题意：

一个n*n的矩阵初始每个元素值为0，左上角行列号最小，右下角行列号最大，两种操作：c x1 y1 x2 y2，将区间内的每个元素取非运算，Q x y表示询问xy点的值。

代码：

//可以画个图看一下，每次更新(x1,y1),(x2+1,y1),(x1,y2+1),(x2+1,y2+1)这4个点的值
//时所有的点都不会被影响。求每个点的sum(并不真实)，奇数是1，偶数是0.
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=;
int A[maxn][maxn];
 int cas,n,m,x1,x2,y1,y2;
int Lowbit(int x){
    return x&(-x);
}
void Add(int x,int y,int val)
{
    for(int i=x;i<=;i+=Lowbit(i)){
        for(int j=y;j<=;j+=Lowbit(j)){
            A[i][j]+=val;
        }
    }
}
int Query(int x,int y)
{
    int s=;
    for(int i=x;i>;i-=Lowbit(i)){
        for(int j=y;j>;j-=Lowbit(j)){
            s+=A[i][j];
        }
    }
    return s;
}
int main()
{
    scanf("%d",&cas);
    while(cas--){
        scanf("%d%d",&n,&m);
        memset(A,,sizeof(A));
        char ch[];
        while(m--){
            scanf("%s",ch);
            if(ch[]=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                Add(x1,y1,);
                Add(x1,y2+,);
                Add(x2+,y2+,);
                Add(x2+,y1,);
            }
            else{
                scanf("%d%d",&x1,&y1);
                int ans=Query(x1,y1);
                //cout<<ans<<endl;
                printf("%d\n",ans&);
            }
        }
        printf("\n");
    }
    return ;
}