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].
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.
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.
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
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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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![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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\n",query(x,y)&1); 52 } 53 } 54 putchar('\n'); 55 } 56 57 return 0; 58 }