POJ 2155-Matrix（二维树状数组-区间修改 单点查询）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 27355		Accepted: 10003

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

题目意思：

N×N的矩形，元素值只能是0/1。

C X1 Y1 X2 Y2表示将左上角(X1,Y1)和右下角(X2,Y2)构成的矩形内的所有元素值取反；

Q X Y表示查询点(X,Y)处的元素值。

解题思路：

二位树状数组。

我们往往单点修改，区间查询，此时update是从小到大，getsum是从大到小；

现在要求区间修改，单点查询，所以update是从大到小，getsum是从小到大。

#include 
#include 
#include 
#include 
using namespace std;
#define MAXN 1100
int c[MAXN][MAXN],n;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int num)//从大到小修改
{
    for(int i=x; i>0; i-=lowbit(i))
        for(int j=y; j>0; j-=lowbit(j))
            c[i][j]+=num;
}
int sum(int x,int y)//从小到大求和
{
    int res=0;
    for(int i=x; i<=n; i+=lowbit(i))
        for(int j=y; j<=n; j+=lowbit(j))
            res+=c[i][j];
    return res;
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("G:/cbx/read.txt","r",stdin);
    //freopen("G:/cbx/out.txt","w",stdout);
#endif
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int t;
        char op[2];
        memset(c,0,sizeof(c));
        scanf("%d%d",&n,&t);
        while(t--)
        {
            scanf("%s",op);
            if(op[0]=='C')
            {
                int x1,y1,x2,y2;
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                update(x1-1,y1-1,1);
                update(x2,y2,1);
                update(x1-1,y2,-1);
                update(x2,y1-1,-1);
            }
            else if(op[0]=='Q')
            {
                int x,y;
                scanf("%d%d",&x,&y);
                printf("%d\n",sum(x,y)%2);
            }
        }
        printf("\n");
    }
    return 0;
}