[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

Sample Output

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

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 int bit[1010][1010];
 5 int n;
 6 int lb(int x){
 7     return x&(-x);
 8 }
 9 int q(int x,int y){
10     int ans=0;
11     while(x){
12         int i=y;
13         while(i){
14             ans+=bit[x][i];
15             i-=lb(i);
16         }
17         x-=lb(x);
18     }
19     return ans%2;
20 }
21 int c(int x,int y){
22     while(x<=n+5){
23         int i=y;
24         while(i<=n+5){
25             bit[x][i]++;
26             i+=lb(i);
27         }
28         x+=lb(x);
29     }
30     return 0;
31 }
32 int main(){
33     int T;
34     scanf("%d",&T);
35     while(T--){
36         int t;
37         scanf("%d %d",&n,&t);
38         memset(bit,0,sizeof(bit));
39         for(int i=1;i<=t;i++){
40             char op=getchar();
41             while(op!='C'&&op!='Q')op=getchar();
42             switch(op){
43                 case 'C':
44                     int x1,y1,x2,y2;
45                     scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
46                     c(x1,y1);
47                     c(x2+1,y1);
48                     c(x1,y2+1);
49                     c(x2+1,y2+1);
50                     break;
51                 case 'Q':
52                     int x,y;
53                     scanf("%d %d",&x,&y);
54                     printf("%d
",q(x,y));
55                     break;
56                 default:
57                     break;
58             }
59         }
60         puts("");
61     }
62     return 0;
63 }

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