poj2155--Matrix（二维树状数组）

Matrix

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 18021 Accepted: 6755

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

POJ Monthly,Lou Tiancheng

二维的树状数组，一般的修改和查询

对二维的树状数组的修改

void add(int i,int j,int d,int n)
{
    int x , y ;
    for(x = i ; x <= n ; x += lowbit(x))
        for(y = j ; y <= n ; y += lowbit(y))
            c[x][y] += d;
}

查询

int sum(int i,int j)
{
    int a = 0 , x , y ;
    for(x = i ; x >= 0 ; x -= lowbit(x))
        for(y = j ; y >= 0 ; y -= lowbit(y))
            a += c[x][y] ;
    return a ;
}

题意：给出n*n的矩阵，矩阵中的值只能为0或者1，初始值全部是0，C x1 y1 x2 y2 表示在（x1,y1）和(x2,y2)围成的矩形中的所有值变换 ；Q x y 查询该点当前的值，也可以认为是统计该点经过了几次的变化；

在这个题中树状数组由后向前更新，每一个点的值表示由该点向前的所有点变换的次数

对于C的操作

首先 将(1,1)到（x2,y2）的所有点+1 ，代表有该点向前的矩形中的变化次数均+1；再由 (x1-1,y2)(x2,y1-1)均-1，(x1-1,y1-1)+1，平衡掉多操作的点，那么当计算该点变换次数时，由该点向后累加，得到的总和就是该点的变换次数，如果是奇数代表1，否则是0.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int c[1010][1010] ;int lowbit(int x){    return x & -x ;}void add(int i,int j,int d){    int x , y ;    for(x = i ; x > 0 ; x -= lowbit(x))        for(y = j ; y > 0 ; y -= lowbit(y))            c[x][y] += d;}int sum(int i,int j,int n){    int a = 0 , x , y ;    for(x = i ; x <= n ; x += lowbit(x))        for(y = j ; y <= n ; y += lowbit(y))            a += c[x][y] ;    return a ;}int main(){    int t , tt , i , j , n , m , x1 , y1 , x2 , y2 ;    char ch ;    scanf("%d", &t);    for(tt = 1 ; tt <= t ; tt++)    {        scanf("%d %d", &n, &m);        memset(c,0,sizeof(c));        while(m--)        {            getchar();            scanf("%c", &ch);            if( ch == 'C' )            {                scanf("%d %d %d %d", &x1, &y1, &x2, &y2);                add(x2,y2,1);                add(x1-1,y2,-1);                add(x2,y1-1,-1);                add(x1-1,y1-1,1);            }            else            {                scanf("%d %d", &x1, &y1);                printf("%d\n", sum(x1,y1,n)%2 );            }        }        if(tt != t)            printf("\n");    }    return 0;}