POJ 2155 Matrix(二维树状数组，绝对具体)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 20599		Accepted: 7673

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：给出矩阵左上角和右下角坐标，矩阵里的元素 1变0 ，0 变1，然后给出询问，问某个点是多少。

题解：纠结了好久，看了这篇博客后秒懂http://blog.sina.com.cn/s/blog_626489680100k75p.html

先举个一维的样例：你要使区间[x,y]所有加上一个值v,结合树状数组的功能，能够类似扫气球那样，在x处加v, y+1处减1

这样假设你要求x处的值，就转换成求[1,x]的和了，比如 ：一个n=6的数组，一開始为0  0  0   0  0  0

在[2,4]加上2后变成   0   2   0   0   -2    0  这样前缀和 sum[1]=0;sum[2]=2;sum[3]=2;sum[4]=2;sum[5]=0;sum[6]=0;

依次代表了每一个数的值。

二维的也一样，由于二维树状数组的getsum(int x,int y)函数是求矩阵（1,1）~(x,y)的值得和。也就类似于前缀和。原理和一维的一样

仅仅只是线操作改成了平面操作。自己能够画个图感受下。

即：

add(x,y,1);

                add(x,y1+1,-1);

                add(x1+1,y,-1);

                add(x1+1,y1+1,1);

然后查询单点就是求和了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cmath>
#define N 1040
#define ll long long

using namespace std;

int n;

int bit[N][N];

int sum(int i,int j) {
    int s=0;
    while(i>0) {
        int jj=j;
        while(jj>0) {
            s+=bit[i][jj];
            jj-=jj&-jj;
        }
        i-=i&-i;
    }
    return s;
}

void add(int i,int j,int x) {
    while(i<=n) {
        int jj=j;
        while(jj<=n) {
            bit[i][jj]+=x;
            jj+=jj&-jj;
        }
        i+=i&-i;
    }
}

int main() {
    freopen("test.in","r",stdin);
    int t;
    cin>>t;
    while(t--) {
        int q;
        scanf("%d%d ",&n,&q);
        memset(bit,0,sizeof bit);
        char c;
        int x,y,x1,y1;
        while(q--) {
            scanf("%c",&c);
            if(c=='C') {
                scanf("%d%d%d%d",&x,&y,&x1,&y1);
                add(x,y,1);
                add(x,y1+1,-1);
                add(x1+1,y,-1);
                add(x1+1,y1+1,1);//重叠的部分加上
            } else {
                scanf("%d%d",&x,&y);
                printf("%d\n",sum(x,y)%2);
            }
            getchar();
        }
        if(t)printf("\n");
    }
    return 0;
}