POJ - 2155 Matrix 二维树状数组入门
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
题意:一个n*n区域,m次操作 C 代表我们将区域转换状态,Q 为查询当前这个点状态
思路:我们累加转换次数 偶数就还是为0 奇数为1 关键在于我们如何转换x,y,x1,y1这个区域。
我们在翻转x,y,x1,y1的过程中我们也将x1,y1,n,n翻转了我们只需要将其他区域翻转回来即可 详细见代码
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<cstring> #include<ctime> #include<cstdlib> #include<algorithm> using namespace std; #define pb push_back #define pll pair<int,int> #define mp make_pair #define ll long long const int maxn = 1000+10; int tree[maxn][maxn]; int n; int lowbit(int x) { return x&(-x); } void add(int x,int y) { for(int i=x; i<=n; i+=lowbit(i)) for(int j=y; j<=n; j+=lowbit(j)) tree[i][j]++; } int query(int x,int y) { int ans=0; for(int i=x; i>0; i-=lowbit(i)) for(int j=y; j>0; j-=lowbit(j)) ans+=tree[i][j]; return ans; } int main() { int t,m,x1,x2,y1,y2; scanf("%d",&t); while(t--) { char ch[10]; memset(tree,0,sizeof(tree)); scanf("%d %d",&n,&m); while(m--) { scanf("%s",ch); if(ch[0]=='C') { scanf("%d %d %d %d",&x1,&y1,&x2,&y2); add(x1,y1); add(x1,y2+1); add(x2+1,y1); add(x2+1,y2+1); } else { scanf("%d %d",&x1,&y1); printf("%d\n",query(x1,y1)&1); } } printf("\n"); } }
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