POJ·2155·Matrix

Matrix

Time Limit: 3000MS

Memory Limit: 65536K

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

Sample Output

Source

POJ Monthly,Lou Tiancheng

楼教主的题，官方题解似乎是二维树状数组，过段时间再来写。

这是偶第一棵二维线段树啊，之前想了一个小时没想明白为什么queryx的时候要queryy，后来发现这和一维是一个道理，一维要访问所有覆盖某个点的线段，二维则要访问所有覆盖该店的矩形。

Codes:

 1 #include<set>
 2 #include<queue>
 3 #include<vector>
 4 #include<cstdio>
 5 #include<cstdlib>
 6 #include<cstring>
 7 #include<iostream>
 8 #include<algorithm>
 9 using namespace std;
10 const int N = 1010;
11 #define Ch1 (i<<1)
12 #define Ch2 (Ch1|1)
13 #define mid(l,r) ((l+r)>>1)
14 #define For(i,n) for(int i=1;i<=n;i++)
15 #define Rep(i,l,r) for(int i=l;i<=r;i++)
16 
17 struct tnodey{
18     short l,r,mid;
19     int change;
20 };
21 
22 struct tnodex{
23     short l,r,mid;
24     tnodey y[N<<2];
25 }T[N<<2];
26 int X,n,m,x1,y1,x,y,x2,y2;
27 char op;
28 
29 void Buildy(int root,int l,int r,int i){
30     T[root].y[i].l = l; T[root].y[i].r = r; T[root].y[i].mid = mid(l,r);
31     T[root].y[i].change = 0;
32     if(l==r) return;
33     Buildy(root,l,mid(l,r),Ch1); Buildy(root,mid(l,r)+1,r,Ch2);
34 }
35 
36 void Buildx(int i,int l,int r){
37     Buildy(i,1,n,1);
38     T[i].l = l; T[i].r = r; T[i].mid = mid(l,r);
39     if(l==r) return;
40     Buildx(Ch1,l,mid(l,r));Buildx(Ch2,mid(l,r)+1,r);
41 }
42 
43 void Modifyy(int root,int l,int r,int i){
44     if(l==T[root].y[i].l&&T[root].y[i].r==r){
45         T[root].y[i].change ^= 1;
46         return;
47     }
48     if(r<=T[root].y[i].mid) Modifyy(root,l,r,Ch1);else
49     if(l>T[root].y[i].mid)  Modifyy(root,l,r,Ch2);else
50     Modifyy(root,l,T[root].y[i].mid,Ch1),Modifyy(root,T[root].y[i].mid+1,r,Ch2);
51 }
52 
53 void Modifyx(int i,int l,int r){
54     if(l==T[i].l&&T[i].r==r){
55         Modifyy(i,y1,y2,1);
56         return;
57     }
58     if(r<=T[i].mid) Modifyx(Ch1,l,r);else
59     if(l>T[i].mid)  Modifyx(Ch2,l,r);else
60     Modifyx(Ch1,l,T[i].mid),Modifyx(Ch2,T[i].mid+1,r);
61 }
62 
63 int queryy(int root,int i,int x){
64     if(T[root].y[i].l==T[root].y[i].r) return T[root].y[i].change;
65     if(x<=T[root].y[i].mid) return queryy(root,Ch1,x) + T[root].y[i].change;else
66     if(x>T[root].y[i].mid)  return queryy(root,Ch2,x) + T[root].y[i].change;
67 }
68 
69 int queryx(int i,int x){
70     if(T[i].l==T[i].r) return queryy(i,1,y);
71     if(x<=T[i].mid)    return queryx(Ch1,x) + queryy(i,1,y);else
72     if(x>T[i].mid)     return queryx(Ch2,x) + queryy(i,1,y);
73 } 
74 
75 
76 void init(){
77     scanf("%d%d",&n,&m);
78     Buildx(1,1,n);
79     For(i,m){
80         scanf("
");scanf("%c",&op);
81         if(op=='C') {
82             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
83             Modifyx(1,x1,x2);
84         }
85         else        {
86             scanf("%d%d",&x,&y);
87             printf("%d
",queryx(1,x)%2);
88         }
89     }
90     printf("
");
91 }
92 
93 int main(){
94     scanf("%d",&X);
95     For(i,X) init();
96     return 0;
97 }