集训第二十四天（2017/8/23）：二维树状数组&三维树状数组

      今上午解决了昨天遗留的问题——一道关于区间修改，区间查询的树状数组的题目，专门写了一篇结题报告。

     今下午主要做了三维树状数组和二维树状数组的题目，感觉还是有模板的，做这类的题目还是有规律可循的。

       要解决POJ2155Matrix(二维树状数组)和hdu 3584 Cube(三维树状数组)这样的题目，这两道题目很相似，只不过从

二维升到三维，其实思路是一样的。

      首先看 POJ2155Matrix(二维树状数组)：

      

Problem 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. <br> <br>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. <br>

 

Output

For each querying output one line, which has an integer representing A[x, y]. <br> <br>There is a blank line between every two continuous test cases. <br>

 

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

   AC代码：

   #include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

#define maxn 1005

int c[maxn][maxn];

int X,n,t;

int lowbit(int x)

{

    return x&(-x);

}

int sum(int x,int y)

{

    int res=0;

    for(int i=x;i>0;i-=lowbit(i))

       for(int j=y;j>0;j-=lowbit(j))

          res+=c[i][j];

    return res;

}

void add(int x,int y)

{

    for(int i=x;i<=n;i+=lowbit(i))

      for(int j=y;j<=n;j+=lowbit(j))

        c[i][j]+=1;

}

int main()

{   char str;

    int x1,y1,x2,y2;

    scanf("%d",&X);

    while(X--)

    {   memset(c,0,sizeof(c));

        scanf("%d%d",&n,&t);

        while(t--)

        {   getchar();

            scanf("%s",&str);

            if(str=='C')

            {

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

                add(x1,y1);

                add(x2+1,y1);

                add(x1,y2+1);

                add(x2+1,y2+1);

            }

            else

            {

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

                printf("%d\n",sum(x1,y1)%2);

            }

        }

        if(X) printf("\n");

    }

    return 0;

}

    然后是hdu 3584 Cube(三维树状数组)：

   

Problem Description

Given an N*N*N cube A, whose elements are either 0 or 1. A[i, j, k] means the number in the i-th row , j-th column and k-th layer. Initially we have A[i, j, k] = 0 (1 <= i, j, k <= N). 

We define two operations, 1: “Not” operation that we change the A[i, j, k]=!A[i, j, k]. that means we change A[i, j, k] from 0->1,or 1->0. (x1<=i<=x2,y1<=j<=y2,z1<=k<=z2).

0: “Query” operation we want to get the value of A[i, j, k].

 

Input

Multi-cases. First line contains N and M, M lines follow indicating the operation below. Each operation contains an X, the type of operation. 1: “Not” operation and 0: “Query” operation. If X is 1, following x1, y1, z1, x2, y2, z2. If X is 0, following x, y,
z.

 

Output

For each query output A[x, y, z] in one line. (1<=n<=100 sum of m <=10000)

 

Sample Input

2 5
1 1 1 1 1 1 1
0 1 1 1
1 1 1 1 2 2 2
0 1 1 1
0 2 2 2

 

Sample Output

1
0
1

 

   /*这题为啥用树状数组呢？

首先1<=N<=100，如果暴力修改的话，最糟糕的时间为O(N*N*N)，然后有M个命令，1<=M<=10000,总耗时为O(M*N*N*N)，也就是100*100*100*10000，而限时1s，神仙难救~

如果把修改次数看成前缀和，修改次数为奇数，答案则为1，修改次数为偶数，则为0，总的耗时为Q(M*logN*logN*logN)，随意可过~

*/

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

#define N 105

int c

,n,m;

int lowbit(int x)

{

    return x&(-x);

}

int sum(int x,int y,int z)

{

    int res=0;

    for(int i=x;i>0;i-=lowbit(i))

      for(int j=y;j>0;j-=lowbit(j))

        for(int k=z;k>0;k-=lowbit(k))

           res+=c[i][j][k];

    return res;

}

void add(int x,int y,int z,int d)

{

    for(int i=x;i<=n;i+=lowbit(i))

      for(int j=y;j<=n;j+=lowbit(j))

        for(int k=z;k<=n;k+=lowbit(k))

            c[i][j][k]+=d;

}

int main()

{   int q,x1,y1,z1,x2,y2,z2;

    while(~scanf("%d%d",&n,&m))

    {

        memset(c,0,sizeof(c));

        while(m--)

        {

            scanf("%d",&q);

            if(q==1)

            {

                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);

                add(x1,y1,z1,1);

                add(x2+1,y1,z1,1);

                add(x1,y2+1,z1,1);

                add(x1,y1,z2+1,1);

                add(x2+1,y2+1,z1,1);

                add(x2+1,y1,z2+1,1);

                add(x1,y2+1,z2+1,1);

                add(x2+1,y2+1,z2+1,1);

            }

            else

            {

                scanf("%d%d%d",&x1,&y1,&z1);

                printf("%d\n",sum(x1,y1,z1)%2);

            }

        }

    }

    return 0;

}

   要理解这类多维树状数组0,1二进制计数问题，首先要理解一维树状数组，有兴趣的童鞋请看 https://wenku.baidu.com/view/1e51750abb68a98271fefaa8.html  ，写得很详细。