【POJ2155】【二维树状数组】Matrix

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

【分析】

算是真正明白二维树状数组了。

维护的时候只要更改矩阵的四个端点就行了。呵呵呵..

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <algorithm>

 4 #include <cstring>

 5 #include <vector>

 6 #include <utility>

 7 #include <iomanip>

 8 #include <string>

 9 #include <cmath>

10 #include <map>

11 

12 const int MAXN = 1000 + 10; 

13 const int MAX = 32000 + 10; 

14 using namespace std;

15 int n, m;//m为操作次数 

16 int C[MAXN][MAXN];

17 

18 int lowbit(int x){return x&-x;}

19 /*int sum(int x, int y){

20     int cnt = 0, tmp;

21     while (x > 0){

22           tmp = y;

23           while (tmp > 0){

24                 cnt += C[x][tmp];

25                 tmp -= lowbit(tmp);

26           }

27           x -= lowbit(x);

28     } 

29     return cnt;

30 }

31 void add(int x, int y, int val){

32      int tmp;

33      while (x <= 1000){

34            tmp = y;

35            while (tmp <= 1000){

36                  C[x][tmp] += val;

37                  tmp += lowbit(tmp);

38            }

39            x += lowbit(x);

40      }

41      return;

42 }*/

43 void add(int x,int y)  {  

44     int i,k;  

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

46         for(k=y; k<=n; k+=lowbit(k))  

47             C[i][k]++;  

48 }  

49 int sum(int x,int y)  {  

50     int i,k,cnt = 0;  

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

52         for(k=y; k>0; k-=lowbit(k))  

53             cnt += C[i][k];  

54     return cnt;  

55 }  

56 

57 void work(){

58      scanf("%d%d", &n, &m);

59      for (int i = 1; i <= m; i++){

60          char str[2];

61          scanf("%s", str);

62          if (str[0] == 'Q'){

63             int x, y;

64             scanf("%d%d", &x, &y);

65             //x++;y++;

66             printf("%d\n", sum(x, y)%2);

67          }else if (str[0] == 'C'){

68                int x1, y1, x2, y2;

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

70                x1++;y1++;x2++;y2++;

71                add(x2, y2);

72                add(x2, y1 - 1);

73                add(x1 - 1, y2);

74                add(x1 - 1, y1 - 1);

75          }

76      }

77 }

78 

79 int main(){

80     int T;

81     #ifdef LOCAL

82     freopen("data.txt",  "r",  stdin);

83     freopen("out.txt",  "w",  stdout); 

84     #endif

85     scanf("%d", &T);

86     while (T--){ 

87           memset(C, 0, sizeof(C));

88           work();

89           printf("\n");

90     }

91     return 0;

92 }

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