Matrix
Time Limit: 3000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2155
64-bit integer IO format: %lld Java class name: Main
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
解题：二维线段树，第一次玩这鬼玩意，调试了N久。。。挫。。。。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <cmath>
5 #include <algorithm>
6 #include <climits>
7 #include <vector>
8 #include <queue>
9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1001;
18 struct subtree{
19 int lt,rt;
20 bool val;
21 };
22 struct node{
23 int lt,rt;
24 subtree sb[maxn<<2];
25 };
26 node tree[maxn<<2];
27 int n,m,ans;
28 void sub(int lt,int rt,int v,subtree *sb){
29 sb[v].lt = lt;
30 sb[v].rt = rt;
31 sb[v].val = false;
32 if(lt == rt) return;
33 int mid = (lt + rt)>>1;
34 sub(lt,mid,v<<1,sb);
35 sub(mid+1,rt,v<<1|1,sb);
36 }
37 void build(int lt,int rt,int v){
38 tree[v].lt = lt;
39 tree[v].rt = rt;
40 sub(1,n,1,tree[v].sb);
41 if(lt == rt) return;
42 int mid = (lt + rt)>>1;
43 build(lt,mid,v<<1);
44 build(mid+1,rt,v<<1|1);
45 }
46 void subupdate(int lt,int rt,int v,subtree *sb){
47 if(sb[v].lt >= lt && sb[v].rt <= rt){
48 sb[v].val ^= 1;
49 return;
50 }
51 if(lt <= tree[v<<1].rt) subupdate(lt,rt,v<<1,sb);
52 if(rt >= tree[v<<1|1].lt) subupdate(lt,rt,v<<1|1,sb);
53 }
54 void update(int lt,int rt,int y1,int y2,int v){
55 if(tree[v].lt >= lt && tree[v].rt <= rt){
56 subupdate(y1,y2,1,tree[v].sb);
57 return;
58 }
59 if(lt <= tree[v<<1].rt) update(lt,rt,y1,y2,v<<1);
60 if(rt >= tree[v<<1|1].lt) update(lt,rt,y1,y2,v<<1|1);
61
62 }
63 void subquery(int y,int v,subtree *sb){
64 ans ^= sb[v].val;
65 if(sb[v].lt == sb[v].rt) return;
66 if(y <= sb[v<<1].rt) subquery(y,v<<1,sb);
67 else subquery(y,v<<1|1,sb);
68 }
69 void query(int x,int y,int v){
70 subquery(y,1,tree[v].sb);
71 if(tree[v].lt == tree[v].rt) return;
72 if(x <= tree[v<<1].rt) query(x,y,v<<1);
73 else query(x,y,v<<1|1);
74 }
75 int main(){
76 int t,x1,y1,x2,y2;
77 char s[5];
78 scanf("%d",&t);
79 while(t--){
80 scanf("%d %d",&n,&m);
81 build(1,n,1);
82 for(int i = 0; i < m; ++i){
83 scanf("%s",s);
84 if(s[0] == ‘Q‘){
85 ans = 0;
86 scanf("%d %d",&x1,&y1);
87 query(x1,y1,1);
88 printf("%d\n",ans);
89 }else if(s[0] == ‘C‘){
90 scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
91 update(x1,x2,y1,y2,1);
92 }
93 }
94 puts("");
95 }
96 return 0;
97 }