POJ2155 Matrix

                                                                                                               Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12970		Accepted: 4866

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

最近开始重新学习树状数组和线段树，希望能有所成长吧。

这题是二维树状数组。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1005;
int tree[maxn][maxn];
int N,T;
inline int Lowbit(int x)
{
    return x&(-x);
}
void Update(int x,int y,int c)
{
    int i,j;
    for(i=x; i<=N; i+=Lowbit(i))
        for(j=y; j<=N; j+=Lowbit(j))
            tree[i][j]+=c;
}
int Getsum(int x,int y)
{
    int i,j;
    int temp=0;
    for(i=x; i>=1; i-=Lowbit(i))
        for(j=y; j>=1; j-=Lowbit(j))
            temp+=tree[i][j];
    return temp;
}
int main()
{
    int X,x1,x2,y1,y2;
    char c;
    scanf("%d",&X);
    while(X--)
    {
        scanf("%d%d",&N,&T);
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
                tree[i][j]=0;
        while(T--)
        {
            getchar();
            scanf("%c",&c);
            if(c=='C')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                x2++;
                y2++;
                Update(x2,y2,1);
                Update(x1,y1,1);
                Update(x2,y1,-1);
                Update(x1,y2,-1);
            }
            else
            {
                scanf("%d%d",&x1,&y1);
                int ans=Getsum(x1,y1);
                printf("%d\n",ans%2);
            }
        }
        printf("\n");
    }
    return 0;
}