【POJ2155】【二维树状数组】Matrix

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

【分析】

算是真正明白二维树状数组了。

维护的时候只要更改矩阵的四个端点就行了。呵呵呵..

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <map>

 const int MAXN =  + ;
 const int MAX =  + ;
 using namespace std;
 int n, m;//m为操作次数
 int C[MAXN][MAXN];

 int lowbit(int x){return x&-x;}
 /*int sum(int x, int y){
     int cnt = 0, tmp;
     while (x > 0){
           tmp = y;
           while (tmp > 0){
                 cnt += C[x][tmp];
                 tmp -= lowbit(tmp);
           }
           x -= lowbit(x);
     }
     return cnt;
 }
 void add(int x, int y, int val){
      int tmp;
      while (x <= 1000){
            tmp = y;
            while (tmp <= 1000){
                  C[x][tmp] += val;
                  tmp += lowbit(tmp);
            }
            x += lowbit(x);
      }
      return;
 }*/
 void add(int x,int y)  {
     int i,k;
     for(i=x; i<=n; i+=lowbit(i))
         for(k=y; k<=n; k+=lowbit(k))
             C[i][k]++;
 }
 int sum(int x,int y)  {
     int i,k,cnt = ;
     for(i=x; i>; i-=lowbit(i))
         for(k=y; k>; k-=lowbit(k))
             cnt += C[i][k];
     return cnt;
 }  

 void work(){
      scanf("%d%d", &n, &m);
      for (int i = ; i <= m; i++){
          char str[];
          scanf("%s", str);
          if (str[] == 'Q'){
             int x, y;
             scanf("%d%d", &x, &y);
             //x++;y++;
             printf("%d\n", sum(x, y)%);
          }else if (str[] == 'C'){
                int x1, y1, x2, y2;
                scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
                x1++;y1++;x2++;y2++;
                add(x2, y2);
                add(x2, y1 - );
                add(x1 - , y2);
                add(x1 - , y1 - );
          }
      }
 }

 int main(){
     int T;
     #ifdef LOCAL
     freopen("data.txt",  "r",  stdin);
     freopen("out.txt",  "w",  stdout);
     #endif
     scanf("%d", &T);
     while (T--){
           memset(C, , sizeof(C));
           work();
           printf("\n");
     }
     return ;
 }