POJ 2155 Matrix

C - Matrix

Time Limit:3000MS    Memory Limit:65536KB    64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2155

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

---

在POJ上看到这是楼教主出的题，读完题后，觉得是一道比较典型的二维树状数组应用。

题目大概意思是说，给一个N*N的矩阵或者叫网格，初始值都是0，然后两种操作，一种是C，把从x1,y1到x2,y2包起来的矩形内的元素值翻转，即0变1，1变0.另一种操作是Q，查询坐标为x,y的元素的值。

第一次二二地写了暴力模拟，果断TLE。。。

第二次尝试使用二维的树状数组，在POJ的Discuss里看到大牛们分享的一种思路。

在C翻转的操作时，每次分别更新4个点到n,n的矩形，从而起到的效果和模拟x1,y1,x2,y2翻转一样（偶次翻转不变）。

树状数组里记录的是某一个点的变换次数总和，当要查询某个点的值的时候，只要getsum一次，就得到这个点所在的所有区间的总变换次数，最后%2就是结果。

最后输出格式里，注意between两块测试要输出一个空行。

#include <stdio.h>
#include <string.h>
#define N 1005
int a[N][N];
int n;
int lowbit(int x)
{
	return x&-x;
}
void modify(int x,int y,int data)
{
	for(int i=x;i<=n;i+=lowbit(i))
		for(int j=y;j<=n;j+=lowbit(j))
		    a[i][j]+=data;
}
int get_sum(int x,int y)
{
	int ret=0;
	for(int i=x;i>0;i-=lowbit(i))
		for(int j=y;j>0;j-=lowbit(j))
		    ret+=a[i][j];
	return ret;
}
int main()
{
        int x;
        scanf("%d",&x);
        while(x--)
        {
                memset(a,0,sizeof(a));
                int t;
                scanf("%d%d",&n,&t);
                char s[20];
                while(t--)
                {
                        scanf("%s",s);
                        if(s[0]=='C')
                        {
                                int x1,y1,x2,y2;
                                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                                modify(x2+1,y2+1,1);
                                modify(x1,y1,1);
                                modify(x1,y2+1,1);
                                modify(x2+1,y1,1);
                        }
                        else
                        {
                                int x,y;
                                scanf("%d%d",&x,&y);
                                printf("%d\n",get_sum(x,y)%2);
                        }
                }
                if(x!=0)
                        printf("\n");
        }
        return 0;
}