POJ 2155 Matrix二维树状数组+YY（区间更新，单点查询）

Posted 2020-12-29 ymzjj

题目链接：http://poj.org/problem?id=2155

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 32950		Accepted: 11943

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

 

题意概括：

有一个初始值为0的N*N的二维矩阵，有T次操作，每次操作有两种选择：

C : 修改以（x1, y1）为左上角（x2, y2）为右下角的矩阵的值（0和1互换）

Q：查询（x, y）的值为 0 或者 为 1；

解题思路：

涉及到多次区间修改和区间查询的优先考虑线段树和树状数组，这道题巧妙之处在于灵活运用树状数组的前缀和，把区间修改转换单点修改，一维需要标记两个点而二维需要标记四个点，一个点用于发挥效果，另外三个点用于消除效果，因为树状数组维护的是前缀和，而对于二维树状数组，查询的则是以（1，1）为左上角，（x，y）为右下角的矩阵的和。

第二就是我们可以借助修改次数的奇偶性来判断该点的值为 0 / 1；

例如：N = 3；C：x1 = 1, y1 = 1, x2 = 2, y2 = 2; 

1		1
	（x，y）
1		1

 

 

 

 

(因为我们只需要知道奇偶性，所以矩阵外的点加1即可消除效果）

 

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #define ll long long int;
 7 #define INF 0x3f3f3f3f
 8 using namespace std;
 9 const int MAXN = 1e3+10;
10 
11 int mmp[MAXN][MAXN];
12 int N, T;
13 
14 int lowbit(int x)
15 {
16     return x&(-x);
17 }
18 
19 void add(int x, int y, int value)
20 {
21     for(int i = x; i <= N; i += lowbit(i))
22     for(int j = y; j <= N; j += lowbit(j))
23         mmp[i][j]+=value;
24 }
25 
26 int sum(int x, int y)
27 {
28     int res = 0;
29     for(int i = x; i > 0; i -= lowbit(i))
30     for(int j = y; j > 0; j -= lowbit(j))
31         res+=mmp[i][j];
32     return res;
33 }
34 
35 void init()
36 {
37     for(int i = 0; i <= N; i++)
38         for(int j = 0; j <= N; j++)
39         mmp[i][j] = 0;
40 }
41 
42 int main()
43 {
44     int T_case;
45     char com[3];
46     int x, y, x1, x2, y1, y2;
47     scanf("%d", &T_case);
48     while(T_case--)
49     {
50         scanf("%d%d", &N, &T);
51         init();
52         while(T--)
53         {
54             scanf("%s", &com);
55             if(com[0] == ‘C‘)
56             {
57                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
58                 add(x1, y1, 1);
59                 add(x2+1, y1, 1);
60                 add(x1, y2+1, 1);
61                 add(x2+1, y2+1, 1);
62             }
63             else if(com[0] == ‘Q‘)
64             {
65                 scanf("%d%d", &x, &y);
66                 int res = 0;
67                 res = sum(x, y);
68                // printf("res: %d
", res);
69                 if(res%2) printf("1
");
70                 else printf("0
");
71             }
72         }
73         puts("");
74     }
75     return 0;
76 }

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