poj 2155 Matrix 二维树状数组

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

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

Sample Output

1001

Source

#include<cstdio>#include<string>#include<iostream>#include<string.h>using namespace std;int c[1005][1005],n;int lowbit(int x){    return x&(-x);}void update(int x,int y,int k ){    for(int i=x; i<=n; i+=lowbit(i))        for(int j=y; j<=n; j+=lowbit(j))        {            c[i][j]+=k;        }}int sum(int x,int y){    int ans=0;    for(int i=x; i>0; i-=lowbit(i))        for(int j=y; j>0; j-=lowbit(j))            ans+=c[i][j];    return ans;}int main(){    int t,m,x1,x2,y1,y2;    char g[100];    scanf("%d",&t);    while(t--)    {      memset(c,0,sizeof(c));        scanf("%d %d",&n,&m);        for(int i=1; i<=m; i++)        {            scanf("%s",g);            if(g[0]=='C')            {                scanf("%d %d %d %d",&x1,&y1,&x2,&y2);                getchar();                update(x1,y1,1);                update(x1,y2+1,1);                update(x2+1,y1,1);                update(x2+1,y2+1,1);            }            else            {                scanf("%d %d",&x1,&y1);                getchar();                printf("%d\n",sum(x1,y1)%2);            }        }        printf("\n");    }    return 0;}

今天初步学习树状数组，虽然还有很多不懂，但这也是一个进步吧

下面是转载的一位大牛的博客

楼教主出的二维树状数组。

 

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。

 

纠结好久了，一直没什么好思路，看discuss说四个角神马的，我搜了下，理解了，树状数组里记录该点的变幻次数，或者直接%2也行。

 

查询的时候Getsum得到的是该点在所有区间的总变幻次数，最后%2就是结果。

 

建图的时候死活想不通，杂四个点的坐标是那个 = =。。。刚才协会开会了，在路上想通了，我想的0,0坐标是类似坐标轴的那种，在左下角。。。而矩阵的0 0 应该是在左上角。。这样，什么都通了 = =。。。

 

还有，更新的时候有的减1了，死循环了，发现错误了，树状数组里是不能使用下标为1的，所以更新的时候把下标都加一即可。

 

实验了下，%2和&1速度一样。。。

 

提供坐标的图。。

 

我见另一种做法，见 http://3214668848.blog.163.com/blog/static/48764919201052484413539/

 

就是原来的Updata改成Getsum。。。Getsum改成Updata。只不过每个坐标不用++了。

 

 

 

这题我还是有点迷茫的。刚才上机组课想了下，经过我画图以及认真分析（//害羞）代码，终于理解了。。。继续画图先。。

 

 

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 1010;int c[MAX][MAX];int n;int Lowbit(int x){return x & (-x);}void Updata(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 ncases,m;int x1,y1,x2,y2;char ch[2];scanf("%d",&ncases);while( ncases-- ){memset(c,0,sizeof(c));scanf("%d%d",&n,&m);while( m-- ){scanf("%s",ch);if( ch[0] == 'C' ){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);x1++; y1++; x2++; y2++;Updata(x2,y2);Updata(x1-1,y1-1);Updata(x1-1,y2);Updata(x2,y1-1);}else{scanf("%d%d",&x1,&y1);printf("%d/n",Get(x1,y1)%2);}}printf("/n");}return 0;}