Matrix+POJ+二维树状数组初步

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18387		Accepted: 6919

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

解决方案：直接用二维树状数组来解。

code:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1003;
int C[maxn][maxn];
int N,Q;
void update(int x,int y,int v){

  for(int i=x;i<=N;i+=(i&-i)){
    for(int j=y;j<=N;j+=(j&-j)){
        C[i][j]+=v;
    }
  }

}///二维树状数组值更新块矩阵
int sum(int x,int y){
   int sum=0;
   for(int i=x;i>0;i-=(i&-i)){
    for(int j=y;j>0;j-=(j&-j)){
       sum+=C[i][j];
    }
   }
   return sum;
}///二维树状数组值求一个元素的值
int main()
{   int t;
    char op;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&N,&Q);
        memset(C,0,sizeof(C));
        for(int i=0;i<Q;i++){
          scanf("\n%c",&op);
          if(op=='C'){
            int x1,y1,x2,y2;
           // cout<<"sdff";
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            x2++,y2++;
            update(x2,y2,1);
            update(x1,y2,-1);
            update(x2,y1,-1);
            update(x1,y1,1);
          }
          else{
            int x,y;
            scanf("%d%d",&x,&y);
            printf("%d\n",sum(x,y)%2);
          }

        }
        if(t) printf("\n");
    }
    return 0;
}