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

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #define maxn 1005

 5 #define lowbit(x) ((x)&(-x))

 6 int aa[maxn][maxn];

 7 int nn;

 8 void ope(int x ,int y ,int val)

 9  {

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

11    {

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

13     {

14       aa[i][j]+=val;

15     }

16    }

17  }

18  int clac(int  x,int y)

19  {

20    int ans=0;

21    for(int i=x;i<=nn ;i+=lowbit(i))

22    {

23      for(int j=y ;j<=nn ;j+=lowbit(j))

24      {

25        ans+=aa[i][j];

26      }

27    }

28    return ans;

29  }

30 struct node

31  {

32    int x;

33    int y;

34  };

35 

36 int main()

37 {

38     int tt,xx;

39     char str[5];

40     node sa,sb;

41     scanf("%d",&xx);

42     while(xx--)

43     {

44      memset(aa,0,sizeof(aa));

45      scanf("%d%d",&nn,&tt);

46      while(tt--)

47      {

48      scanf("%s",&str);

49      if(str[0]=='C')

50      {

51       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);

52       sa.x--; //左上角全体加1

53       sa.y--;

54       ope(sb.x,sb.y,1);

55       ope(sa.x,sb.y,-1);

56       ope(sb.x,sa.y,-1); 

57       ope(sa.x,sa.y,1);

58      }

59     else

60     {

61      scanf("%d%d",&sa.x,&sa.y);

62      printf("%d\n",clac(sa.x,sa.y)&1);

63     }

64    }

65    printf("\n");

66  }

67   return 0;

68 }

View Code

改进版..
代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #define maxn 1005

 5 #define lowbit(x) ((x)&(-x))

 6 int aa[maxn][maxn];

 7 int nn;

 8 void ope(int x ,int y )

 9  {

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

11    {

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

13     {

14       aa[i][j]=aa[i][j]^1;

15     }

16    }

17  }

18  int clac(int  x,int y)

19  {

20    int ans=0;

21    for(int i=x;i<=nn ;i+=lowbit(i))

22    {

23      for(int j=y ;j<=nn ;j+=lowbit(j))

24      {

25        ans+=aa[i][j];

26      }

27    }

28    return ans;

29  }

30 struct node

31  {

32    int x;

33    int y;

34  };

35 

36 int main()

37 {

38     int tt,xx;

39     char str[5];

40     node sa,sb;

41     scanf("%d",&xx);

42     while(xx--)

43     {

44      memset(aa,0,sizeof(aa));

45      scanf("%d%d",&nn,&tt);

46      while(tt--)

47      {

48      scanf("%s",&str);

49      if(str[0]=='C')

50      {

51       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);

52       sa.x--; //左上角全体加1

53       sa.y--;

54       ope(sb.x,sb.y);

55       ope(sa.x,sb.y);

56       ope(sb.x,sa.y); 

57       ope(sa.x,sa.y);

58      }

59     else

60     {

61      scanf("%d%d",&sa.x,&sa.y);

62      printf("%d\n",clac(sa.x,sa.y)&1);

63     }

64    }

65    printf("\n");

66  }

67   return 0;

68 }

View Code

 采用树状数组第一种方法

传统的方法：

代码：435ms

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<stdlib.h>

 4 #define maxn 1005

 5 #define lowbit(x) ((x)&(-x))

 6 int aa[maxn][maxn];

 7 int nn;

 8 void ope(int x ,int y )

 9  {

10    for(int i=x ;i<=nn ;i+=lowbit(i))

11      for(int j=y ;j<=nn ;j+=lowbit(j))

12          aa[i][j]=aa[i][j]^1;

13  }

14  int clac(int  x,int y)

15  {

16    int ans=0,i,j;

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

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

19        ans+=aa[i][j];

20    return ans;

21  }

22 struct node

23  {

24    int x,y;

25  };

26 int main()

27 {

28     int tt,xx;

29     char str[5];

30     node sa,sb;

31     scanf("%d",&xx);

32     while(xx--)

33     {

34      memset(aa,0,sizeof(aa));

35      scanf("%d%d",&nn,&tt);

36      while(tt--)

37      {

38      scanf("%s",&str);

39      if(str[0]=='C')

40      {

41       scanf("%d%d%d%d",&sa.x,&sa.y,&sb.x,&sb.y);

42        sb.x++; //左上角全体加1

43        sb.y++;

44       ope(sb.x,sb.y);

45       ope(sa.x,sb.y);

46       ope(sb.x,sa.y);

47       ope(sa.x,sa.y);

48      }

49     else

50     {

51      scanf("%d%d",&sa.x,&sa.y);

52      printf("%d\n",clac(sa.x,sa.y)&1);

53     }

54    }

55    printf("\n");

56  }

57   return 0;

58 }

View Code