POJ 2155 Matrix （二维树状数组）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17224		Accepted: 6460

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

很裸的题。

可以使用二维树状数组。

二维的写起来很方便，两重循环。

如果是要修改(x1,y1)  -  (x2,y2)的矩形区域。

那么可以在(x1,y1) 出加1，在(x2+1,y1)处加1，在(x1,y2+1)处加1，在(x2+1,y2+1)处加1 。。

画个图就知道了，查询单点就是求和。

/* ***********************************************
Author        :kuangbin
Created Time  :2014/5/23 22:34:04
File Name     :E:2014ACM专题学习数据结构二维树状数组POJ2155.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 1010;
int lowbit(int x)
{
    return x&(-x);
}
int c[MAXN][MAXN];
int n;
int sum(int x,int y)
{
    int ret = 0;
    for(int i = x;i > 0;i -= lowbit(i))
        for(int j = y;j > 0;j -= lowbit(j))
            ret += c[i][j];
    return ret;
}
void add(int x,int y,int val)
{
    for(int i = x;i <= n;i += lowbit(i))
        for(int j = y;j <= n;j += lowbit(j))
            c[i][j] += val;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int q;
        scanf("%d%d",&n,&q);
        memset(c,0,sizeof(c));
        char op[10];
        int x1,y1,x2,y2;
        while(q--)
        {
            scanf("%s",op);
            if(op[0] == 'C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                add(x1,y1,1);
                add(x2+1,y1,1);
                add(x1,y2+1,1);
                add(x2+1,y2+1,1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                if(sum(x1,y1)%2 == 0)printf("0
");
                else printf("1
");
            }
        }
        if(T > 0)printf("
");
    }
    return 0;
}