[poj2155]Matrix(二维树状数组)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25004		Accepted: 9261

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

 

继续继续

 

二维树状数组果然比二维线段树简单多了

讲一下二维树状数组

其实我也不清楚多出来的一维怎么做

但既然多套了一重循环就算作是二维了

文字说不清，自己仿照一维画一个图就明白了

再说这道题

假设只有一维，我们可以用树状数组维护一个差分数组，区间首尾打标记+1，求和即可

那么推广到二维，把维护差分数组的方式看成打一个标记

四个点+1，对询问求一遍和模2

尹神的办法zrl说可以推广，而这种办法只对01有效

就是说对于正常的差分数组，区间修改应该是首加尾减

到了二维应该这样维护

-1 +1

+1 -1

就这样吧

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 int bit[][];
 int n;
 int lb(int x){
     return x&(-x);
 }
 int q(int x,int y){
     int ans=;
     while(x){
         int i=y;
         while(i){
             ans+=bit[x][i];
             i-=lb(i);
         }
         x-=lb(x);
     }
     return ans%;
 }
 int c(int x,int y){
     while(x<=n+){
         int i=y;
         while(i<=n+){
             bit[x][i]++;
             i+=lb(i);
         }
         x+=lb(x);
     }
     return ;
 }
 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         int t;
         scanf("%d %d",&n,&t);
         memset(bit,,sizeof(bit));
         for(int i=;i<=t;i++){
             char op=getchar();
             while(op!='C'&&op!='Q')op=getchar();
             switch(op){
                 case 'C':
                     int x1,y1,x2,y2;
                     scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                     c(x1,y1);
                     c(x2+,y1);
                     c(x1,y2+);
                     c(x2+,y2+);
                     break;
                 case 'Q':
                     int x,y;
                     scanf("%d %d",&x,&y);
                     printf("%d\n",q(x,y));
                     break;
                 default:
                     break;
             }
         }
         puts("");
     }
     return ;
 }