POJ2155 二维树状数组

Matrix

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

1
0
0
1

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二维的树状数组，二维求和。
代码很简单：

 1 void add(int x,int y,int v)
 2 {
 3     for(int i=x;i<=n;i+=lowbit(i))
 4     for(int j=y;j<=n;j+=lowbit(j))
 5     sz[i][j]+=v;
 6 }
 7 int query(int x,int y)
 8 {
 9     int ans=0;
10     for(int i=x;i>0;i-=lowbit(i))
11     for(int j=y;j>0;j-=lowbit(j))
12     ans+=sz[i][j];
13     return ans;
14 }

代码短，效率高。但是如果错了，着实不好调，不是代码复杂，而是树状数组的树形结构实在不向线段树那样能一眼看明白。

错了就再写一遍吧，反正也不长。

这个题有一点比较坑，就是每一组数据后都要有一个空行。要命了，调了半天，不懂英语要人命啊！


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 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=1005;
 5 int t,n,m;
 6 int sz[maxn][maxn];
 7 char c[2];
 8 int lowbit(int x)
 9 {
10     return x&(-x);
11 }
12 void add(int x,int y,int v)
13 {
14     for(int i=x;i<=n;i+=lowbit(i))
15     for(int j=y;j<=n;j+=lowbit(j))
16     sz[i][j]+=v;
17 }
18 int query(int x,int y)
19 {
20     int ans=0;
21     for(int i=x;i>0;i-=lowbit(i))
22     for(int j=y;j>0;j-=lowbit(j))
23     ans+=sz[i][j];
24     return ans;
25 }
26 int main()
27 {
28     scanf("%d",&t);
29     while(t--)
30     {
31         memset(sz,0,sizeof(sz));
32         scanf("%d%d",&n,&m);
33         for(int i=0;i<m;i++)
34         {
35             
36             scanf("%s",c);
37             if(c[0]=='C')
38             {
39                 int x1,y1,x2,y2;
40                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
41                 ++x2;++y2;
42                 add(x1,y1,1);
43                 add(x2,y2,1);
44                 add(x1,y2,1);
45                 add(x2,y1,1);
46             }
47             else 
48             {
49                 int x,y;
50                 scanf("%d%d",&x,&y);
51                 printf("%d
",query(x,y)&1);
52             }
53         }
54     putchar('
');
55     }
56 
57     return 0;
58 }

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