POJ 2155 Matrix【二維樹狀數組+YY（區間更新，單點查詢）】

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題目連結：http://poj.org/problem?id=2155

Matrix

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

DescriptionGiven 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]. InputThe 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. OutputFor 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

12 10C 2 1 2 2Q 2 2C 2 1 2 1Q 1 1C 1 1 2 1C 1 2 1 2C 1 1 2 2Q 1 1C 1 1 2 1Q 2 1

Sample Output

1001

Source

POJ Monthly,Lou Tiancheng

 

題意概括：

有一個初始值為0的N*N的二維矩陣，有T次操作，每次操作有兩種選擇：

C : 修改以（x1, y1）為左上方（x2, y2）為右下角的矩陣的值（0和1互換）

Q：查詢（x, y）的值為 0 或者 為 1；

解題思路：

涉及到多次區間修改和區間查詢的優先考慮線段樹和樹狀數組，這道題巧妙之處在於靈活運用樹狀數組的首碼和，把區間修改轉換單點修改，一維需要標記兩個點而二維需要標記四個點，一個點用於發揮效果，另外三個點用於消除效果，因為樹狀數組維護的是首碼和，而對於二維樹狀數組，查詢的則是以（1，1）為左上方，（x，y）為右下角的矩陣的和。

第二就是我們可以藉助修改次數的奇偶性來判斷該點的值為 0 / 1；

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

1		1
	（x，y）
1		1

 

 

 

 

(因為我們只需要知道奇偶性，所以矩陣外的點加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\n", res);69                 if(res%2) printf("1\n");70                 else printf("0\n");71             }72         }73         puts("");74     }75     return 0;76 }

View Code

 

POJ 2155 Matrix【二維樹狀數組+YY（區間更新，單點查詢）】