poj2155 树状数组 Matrix

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 14826		Accepted: 5583

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

树状数组好强大，在这一题中，说是原始都是一些０　１，通过操作可以使１变０　０变１，在这一题中，我们可以发现，是一个区间的更新，然后是，求得一个点的值，这和我们一般用法刚好相反，其实，我们可以转换角度，其实，如果，是从最上向下更新，然后从下到上求和，这样，我们不就把一个点的值，转化成了求一个区间的值了么？也就基于这样的思想，我们在实际的用法中，要注意把向下的时候，+１然后，在重叠处－１，画画图就知道了！说也说不清楚！很好的题啊！原本是想弄线段树的，但是有的复杂，还有可以暴内存！

#include<iostream>

#include <string.h>

#include<stdio.h>

using namespace std;

#define MAXN  1005

int n;

int matrix[MAXN][MAXN];

int lowbit(int x)

{

    return x&(-x);

}

int change(int x,int y,int val)//从上到下更新

{

   int i,j;

   for(i=x;i<=n;i=i+lowbit(i))

      for(j=y;j<=n;j=j+lowbit(j))

        {

           matrix[i][j]+=val;

        }

    return 0;

}

int getsum(int x,int y)//从下向上求和

{

    int i,j,re=0;

    for(i=x;i>0;i=i-lowbit(i))

        for(j=y;j>0;j=j-lowbit(j))

        {

            re+=matrix[i][j];

        }

    return re;

}

int main ()

{

    int tcase,ask,x1,x2,y1,y2;

    char c;

    scanf("%d",&tcase);

    while(tcase--)

    {

        memset(matrix,0,sizeof(matrix));

        scanf("%d%d",&n,&ask);

        getchar();

        while(ask--)

        {

            c=getchar();

            if(c=='C')

            {

                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

                x2++;y2++;

                change(x1,y1,1);

                change(x2,y2,1);

                change(x1,y2,-1);

                change(x2,y1,-1);

            }

            else

            {

                scanf("%d%d",&x1,&y1);

                printf("%d\n",1&getsum(x1,y1));//判定奇偶

            }

            getchar();

        }

        printf("\n");



    }

    return 0;

}