POJ 2155 Matrix(二维树状数组)

Matrix Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 28817 Accepted: 10507

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

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

Source

POJ Monthly,Lou Tiancheng

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一维的类似题目

#include <iostream> #include <stdio.h> #include <string.h> #include <math.h> #include <algorithm> using namespace std; #define LL long long const int N = 1000; int n,Q; int sum[N+3][N+3]; 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)) { sum[i][j]++; } } } int query(int x,int y) { int ans=0; for(int i=x;i>=1;i-=lowbit(i)) { for(int j=y;j>=1;j-=lowbit(j)) { ans+=sum[i][j]; } } return ans; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d %d",&n,&Q); memset(sum,0,sizeof sum); while(Q--) { char c; cin>>c; if(c=='C') { int x1,x2,y1,y2; scanf("%d %d %d %d",&x1,&y1,&x2,&y2); add(x1,y1); add(x2+1,y2+1); add(x2+1,y1); add(x1,y2+1); } else { int x,y; scanf("%d %d",&x,&y); int ans=query(x,y); printf("%d\n",ans%2); } } if(T) printf("\n"); } }