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[poj2155]Matrix(二维树状数组)

2024-08-11 10:15:01   218人阅读

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]. 

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

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
 
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
技术分享
 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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[poj2155]Matrix(二维树状数组)


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