【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 = 1000 + 10; 
const int MAX = 32000 + 10; 
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 = 0;  
    for(i=x; i>0; i-=lowbit(i))  
        for(k=y; k>0; k-=lowbit(k))  
            cnt += C[i][k];  
    return cnt;  
}  

void work(){
     scanf("%d%d", &n, &m);
     for (int i = 1; i <= m; i++){
         char str[2];
         scanf("%s", str);
         if (str[0] == 'Q'){
            int x, y;
            scanf("%d%d", &x, &y);
            //x++;y++;
            printf("%d\n", sum(x, y)%2);
         }else if (str[0] == '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 - 1);
               add(x1 - 1, y2);
               add(x1 - 1, y1 - 1);
         }
     }
}

int main(){
    int T;
    #ifdef LOCAL
    freopen("data.txt",  "r",  stdin);
    freopen("out.txt",  "w",  stdout); 
    #endif
    scanf("%d", &T);
    while (T--){ 
          memset(C, 0, sizeof(C));
          work();
          printf("\n");
    }
    return 0;
}