[poj2155]Matrix(二维树状数组)
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Matrix
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 25004 | Accepted: 9261 |
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].
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.
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.
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
继续继续
二维树状数组果然比二维线段树简单多了
讲一下二维树状数组
其实我也不清楚多出来的一维怎么做
但既然多套了一重循环就算作是二维了
文字说不清,自己仿照一维画一个图就明白了
再说这道题
假设只有一维,我们可以用树状数组维护一个差分数组,区间首尾打标记+1,求和即可
那么推广到二维,把维护差分数组的方式看成打一个标记
四个点+1,对询问求一遍和模2
尹神的办法zrl说可以推广,而这种办法只对01有效
就是说对于正常的差分数组,区间修改应该是首加尾减
到了二维应该这样维护
-1 +1
+1 -1
就这样吧
1 #include<stdio.h> 2 #include<stdlib.h> 3 #include<string.h> 4 int bit[1010][1010]; 5 int n; 6 int lb(int x){ 7 return x&(-x); 8 } 9 int q(int x,int y){ 10 int ans=0; 11 while(x){ 12 int i=y; 13 while(i){ 14 ans+=bit[x][i]; 15 i-=lb(i); 16 } 17 x-=lb(x); 18 } 19 return ans%2; 20 } 21 int c(int x,int y){ 22 while(x<=n+5){ 23 int i=y; 24 while(i<=n+5){ 25 bit[x][i]++; 26 i+=lb(i); 27 } 28 x+=lb(x); 29 } 30 return 0; 31 } 32 int main(){ 33 int T; 34 scanf("%d",&T); 35 while(T--){ 36 int t; 37 scanf("%d %d",&n,&t); 38 memset(bit,0,sizeof(bit)); 39 for(int i=1;i<=t;i++){ 40 char op=getchar(); 41 while(op!=\'C\'&&op!=\'Q\')op=getchar(); 42 switch(op){ 43 case \'C\': 44 int x1,y1,x2,y2; 45 scanf("%d %d %d %d",&x1,&y1,&x2,&y2); 46 c(x1,y1); 47 c(x2+1,y1); 48 c(x1,y2+1); 49 c(x2+1,y2+1); 50 break; 51 case \'Q\': 52 int x,y; 53 scanf("%d %d",&x,&y); 54 printf("%d\\n",q(x,y)); 55 break; 56 default: 57 break; 58 } 59 } 60 puts(""); 61 } 62 return 0; 63 }
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