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 = ;
 int lowbit(int x)
 {
     return x&(-x);
 }
 int c[MAXN][MAXN];
 int n;
 int sum(int x,int y)
 {
     int ret = ;
     for(int i = x;i > ;i -= lowbit(i))
         for(int j = y;j > ;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,,sizeof(c));
         char op[];
         int x1,y1,x2,y2;
         while(q--)
         {
             scanf("%s",op);
             if(op[] == 'C')
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 add(x1,y1,);
                 add(x2+,y1,);
                 add(x1,y2+,);
                 add(x2+,y2+,);
             }
             else
             {
                 scanf("%d%d",&x1,&y1);
                 if(sum(x1,y1)% == )printf("0\n");
                 else printf("1\n");
             }
         }
         if(T > )printf("\n");
     }
     return ;
 }