POJ 2155 二维树状数组解题报告

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数据结构——树状数组

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本文介绍了一种使用二维树状数组解决特定矩阵操作问题的方法。该方法通过更新和查询操作实现了对矩阵中元素的有效翻转及状态查询，适用于处理大规模矩阵的变更指令。

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

【解题报告】
二维树状数组。
好写好调

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 1010
#define lowbit(x) (x&(-x)) 

int c[N][N],n,t,cas;

int sum(int x,int y)   
{    
    int ans=0;    
    for(int i=x;i>0;i-=lowbit(i))     
    for(int j=y;j>0;j-=lowbit(j)) ans+=c[i][j];         
    return ans;   
}  
void update(int x,int y,int k)   
{    
    for(int i=x;i<=n;i+=lowbit(i))     
    for(int j=y;j<=n;j+=lowbit(j)) c[i][j]+=k;        
}

int main()   
{        
    for(scanf("%d",&cas);cas;--cas)    
    {     
        memset(c,0,sizeof(c));     
        scanf("%d%d",&n,&t);     
        getchar();     
        for(int i=1;i<=t;++i)     
        {
            char opt[5];       
            int x1,x2,y1,y2,a,b;
            scanf("%s",opt);      
            if(opt[0]=='C')      
            {       
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);      
                update(x1,y1,1);       
                update(x1,y2+1,1);       
                update(x2+1,y1,1);       
                update(x2+1,y2+1,1);           
            }      
            if(opt[0]=='Q')  
            {       
                scanf("%d%d",&a,&b);               
                printf("%d\n",sum(a,b)%2);      
            }     
        }  
        printf("\n");     
    }    
    return 0;   
}