POJ2155 Matrix

Posted 2020-08-09 SilverNebula

 

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25080		Accepted: 9293

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

 

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

 

——by隔壁老司机RLQ

 

 

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 using namespace std;
 8 const int mxn=1200;
 9 int t[mxn][mxn];
10 int n,m;
11 int lowbit(int x){return x&-x;}
12 int an;
13 void add(int x,int y,int v){
14     while(x<=n){
15         int tmp=y;
16         while(tmp<=n){ t[x][tmp]+=v; tmp+=lowbit(tmp);}
17         x+=lowbit(x);
18     }
19 }
20 int sum(int x,int y){
21     int res=0;
22     while(x){
23         int tmp=y;
24         while(tmp){    res+=t[x][tmp];     tmp-=lowbit(tmp);}
25         x-=lowbit(x);
26     }
27     return res;
28 }
29 int ask(int x1,int y1,int x2,int y2){
30     return sum(x2,y2)+sum(x1-1,y1-1)-sum(x2,y1-1)-sum(x1,y2-1);
31 }
32 int main(){
33     int T;
34     scanf("%d",&T);
35     int i,j;
36     char ch[5];
37     int x1,y1,x2,y2;
38     while(T--){
39         memset(t,0,sizeof t);
40         scanf("%d%d",&n,&m);
41         an=n+5;
42         for(i=1;i<=m;i++){
43             scanf("%s",ch);
44             if(ch[0]==‘C‘){
45                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
46                 add(x1,y1,1);
47                 add(x1,y2+1,-1);
48                 add(x2+1,y1,-1);
49                 add(x2+1,y2+1,1);
50             }
51             else{
52                 scanf("%d%d",&x1,&y1);
53                 printf("%d\n",sum(x1,y1)%2);
54             }
55         }
56         printf("\n");
57     }
58     return 0;
59 }