poj----2155 Matrix(二维树状数组第二类)

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 16950		Accepted: 6369

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

 

代码：

采用树状数组第二种方法

采用更新向下，统计向上的方法....楼教主这道题出的还是比较新颖的......

代码：438ms

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #define maxn 1005
 #define lowbit(x) ((x)&(-x))
 int aa[maxn][maxn];
 int nn;
 void ope(int x ,int y ,int val)
  {
    for(int i=x ;i> ;i-=lowbit(i))
    {
     for(int j=y ;j> ;j-=lowbit(j))
     {
       aa[i][j]+=val;
     }
    }
  }
  int clac(int  x,int y)
  {
    int ans=;
    for(int i=x;i<=nn ;i+=lowbit(i))
    {
      for(int j=y ;j<=nn ;j+=lowbit(j))
      {
        ans+=aa[i][j];
      }
    }
    return ans;
  }
 struct node
  {
    int x;
    int y;
  };

 int main()
 {
     int tt,xx;
     char str[];
     node sa,sb;
     scanf("%d",&xx);
     while(xx--)
     {
      memset(aa,,sizeof(aa));
      scanf("%d%d",&nn,&tt);
      while(tt--)
      {
      scanf("%s",&str);
      if(str[]=='C')
      {
       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
       sa.x--; //左上角全体加1
       sa.y--;
       ope(sb.x,sb.y,);
       ope(sa.x,sb.y,-);
       ope(sb.x,sa.y,-);
       ope(sa.x,sa.y,);
      }
     else
     {
      scanf("%d%d",&sa.x,&sa.y);
      printf("%d\n",clac(sa.x,sa.y)&);
     }
    }
    printf("\n");
  }
   return ;
 }

改进版..
代码：

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #define maxn 1005
 #define lowbit(x) ((x)&(-x))
 int aa[maxn][maxn];
 int nn;
 void ope(int x ,int y )
  {
    for(int i=x ;i> ;i-=lowbit(i))
    {
     for(int j=y ;j> ;j-=lowbit(j))
     {
       aa[i][j]=aa[i][j]^;
     }
    }
  }
  int clac(int  x,int y)
  {
    int ans=;
    for(int i=x;i<=nn ;i+=lowbit(i))
    {
      for(int j=y ;j<=nn ;j+=lowbit(j))
      {
        ans+=aa[i][j];
      }
    }
    return ans;
  }
 struct node
  {
    int x;
    int y;
  };

 int main()
 {
     int tt,xx;
     char str[];
     node sa,sb;
     scanf("%d",&xx);
     while(xx--)
     {
      memset(aa,,sizeof(aa));
      scanf("%d%d",&nn,&tt);
      while(tt--)
      {
      scanf("%s",&str);
      if(str[]=='C')
      {
       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
       sa.x--; //左上角全体加1
       sa.y--;
       ope(sb.x,sb.y);
       ope(sa.x,sb.y);
       ope(sb.x,sa.y);
       ope(sa.x,sa.y);
      }
     else
     {
      scanf("%d%d",&sa.x,&sa.y);
      printf("%d\n",clac(sa.x,sa.y)&);
     }
    }
    printf("\n");
  }
   return ;
 }

采用树状数组第一种方法

传统的方法：

代码：435ms

 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #define maxn 1005
 #define lowbit(x) ((x)&(-x))
 int aa[maxn][maxn];
 int nn;
 void ope(int x ,int y )
  {
    for(int i=x ;i<=nn ;i+=lowbit(i))
      for(int j=y ;j<=nn ;j+=lowbit(j))
          aa[i][j]=aa[i][j]^;
  }
  int clac(int  x,int y)
  {
    int ans=,i,j;
    for(i=x;i> ;i-=lowbit(i))
      for(j=y ;j> ;j-=lowbit(j))
        ans+=aa[i][j];
    return ans;
  }
 struct node
  {
    int x,y;
  };
 int main()
 {
     int tt,xx;
     char str[];
     node sa,sb;
     scanf("%d",&xx);
     while(xx--)
     {
      memset(aa,,sizeof(aa));
      scanf("%d%d",&nn,&tt);
      while(tt--)
      {
      scanf("%s",&str);
      if(str[]=='C')
      {
       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);
        sb.x++; //左上角全体加1
        sb.y++;
       ope(sb.x,sb.y);
       ope(sa.x,sb.y);
       ope(sb.x,sa.y);
       ope(sa.x,sa.y);
      }
     else
     {
      scanf("%d%d",&sa.x,&sa.y);
      printf("%d\n",clac(sa.x,sa.y)&);
     }
    }
    printf("\n");
  }
   return ;
 }