poj 2155 Matrix（树状数组）

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Matrix

Time Limit: 3000MS

Memory Limit: 65536K

Total Submissions: 21533

Accepted: 8051

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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

这题是比较经典的二维树状数组，题意是给你个矩阵里面开始全是0，然后给你两种指令：1：‘C x1,y1,x2,y2’就是将左上角为x1,y1,右下角为x2,y2,的这个矩阵内的数字全部翻转，0变1,1变0,；2:'Q x1 y1',输出a[x1][y1]的值。

树状数组里记录该点的变幻次数，或者直接%2也行。查询的时候sum得到的是该点在所有区间的总变幻次数，最后%2就是结果！

这题的巧妙只初在于重叠消元，将要翻转的矩阵的四个角更新一遍就ok了，去掉重叠部分（结果%2），刚好剩下了这个矩阵翻转了！

如图 黑色的四边形为所要变换的子矩形，为了抵消其他点的变化，把不该变化的再次变化即可，记录变化次数%2，这样就可以判断了！

                   updata(x1,y1,1);    ---->>黑+黄+绿+蓝                   updata(x1,y2+1,1);  ---->>黄+蓝                   updata(x2+1,y1,1);  ---->>绿+蓝                   updata(x2+1,y2+1,1);---->>蓝

                   （对应的四次变换）！

代码：

 #include<stdio.h>#include<string.h>#define V 1010int a[V][V];int N,T;int lowbit(int x) {return x&(-x);}void updata(int x,int y,int k)//更新 {int i,j;for(i=x;i<=N;i+=lowbit(i)) for(j=y;j<=N;j+=lowbit(j))  a[i][j]+=k;}int sum(int x,int y)//求所在点的变化次数 {int i,j;int sum=0;for(i=x;i>0;i-=lowbit(i)) for(j=y;j>0;j-=lowbit(j))   sum+=a[i][j];   return sum;}int main(){int n;scanf("%d",&n);while(n--){scanf("%d%d",&N,&T);memset(a,0,sizeof(a));while(T--){getchar();char s;scanf("%c",&s);int x1,y1,x2,y2;if(s=='C'){scanf("%d%d%d%d",&x1,&y1,&x2,&y2);updata(x1,y1,1);//四个矩形更新 updata(x1,y2+1,1);updata(x2+1,y1,1);updata(x2+1,y2+1,1);}else{scanf("%d%d",&x1,&y1);printf("%d\n",sum(x1,y1)%2);}}printf("\n");}return 0;}

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