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

Sample Output

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("%dn",clac(sa.x,sa.y)&1);
63     }
64    }
65    printf("n");
66  }
67   return 0;
68 }

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("%dn",clac(sa.x,sa.y)&1);
63     }
64    }
65    printf("n");
66  }
67   return 0;
68 }

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("%dn",clac(sa.x,sa.y)&1);
53     }
54    }
55    printf("n");
56  }
57   return 0;
58 }