一道二维树状数组题
这题要用差分的思想
将x1y1x2y2区间取反,就将x1,y1;x2+1,y1;x1,y2+1;x2+1,y2+1;加1
查询就直接输出,注意mod2
Matrix
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 34294 | Accepted: 12409 |
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].
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
#include<cstdio> #include<iostream> #include<cstring> using namespace std; const int maxn = 1e3+10; int t,n,k,tree[maxn][maxn]; int lowbit(int x){ return x&(-x); } void add(int x,int y,int delta){ for(int i=x;i<=n;i+=lowbit(i)) for(int j=y;j<=n;j+=lowbit(j)){ tree[i][j]+=delta; } } int getsum(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; } char getc(){ char ch=getchar(); while(ch!='Q'&&ch!='C') ch=getchar(); return ch; } int main(){ scanf("%d",&t); while(t--){ memset(tree,0,sizeof(tree)); scanf("%d%d",&n,&k); for(int i=1;i<=k;i++){ char ch=getc(); if(ch=='C'){ int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(x1,y1,1);add(x1,y2+1,1);add(x2+1,y1,1);add(x2+1,y2+1,1); } else{ int x1,y1; scanf("%d%d",&x1,&y1); printf("%d ",getsum(x1,y1)%2); } } if(t) printf(" ");//题目要求,要加 } return 0; }