poj 2155 Matrix

Matrix
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 25517 Accepted: 9450

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

【分析】
二维树状数组经典模板题

【代码】

//poj 2155 Matrix
#include<iostream>
#include<cstdio>
#include<cstring>
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
int n,t,T;
int c[1005][1005];
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))
        c[i][j]++;
}
int query(int x,int y)
{
    int ans=0;
    for(int i=x;i;i-=lowbit(i))
      for(int j=y;j;j-=lowbit(j))
        ans+=c[i][j];
    return ans%2;
}
int main()
{
    int i,j,k,x1,y1,x2,y2,x,y;
    scanf("%d",&T);
    while(T--)
    {
        M(c);
        scanf("%d%d",&n,&t);
        while(t--)
        {
            char tmp[2];
            scanf("%s",tmp);
            if(tmp[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",&x,&y);
                printf("%d\n",query(x,y));
            }
        }
        printf("\n");
    }
    return 0;