POJ2155 二维树状数组

Matrix

 POJ - 2155 

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

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二维的树状数组，二维求和。
代码很简单：

void add(int x,int y,int v)
{
    for(int i=x;i<=n;i+=lowbit(i))
    for(int j=y;j<=n;j+=lowbit(j))
    sz[i][j]+=v;
}
int query(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+=sz[i][j];
    return ans;
}

代码短，效率高。但是如果错了，着实不好调，不是代码复杂，而是树状数组的树形结构实在不向线段树那样能一眼看明白。

错了就再写一遍吧，反正也不长。

 

这个题有一点比较坑，就是每一组数据后都要有一个空行。要命了，调了半天，不懂英语要人命啊！

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#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1005;
int t,n,m;
int sz[maxn][maxn];
char c[2];
int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int y,int v)
{
    for(int i=x;i<=n;i+=lowbit(i))
    for(int j=y;j<=n;j+=lowbit(j))
    sz[i][j]+=v;
}
int query(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+=sz[i][j];
    return ans;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(sz,0,sizeof(sz));
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++)
        {
            
            scanf("%s",c);
            if(c[0]=='C')
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                ++x2;++y2;
                add(x1,y1,1);
                add(x2,y2,1);
                add(x1,y2,1);
                add(x2,y1,1);
            }
            else 
            {
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",query(x,y)&1);
            }
        }
    putchar('\n');
    }

    return 0;
}