HDU 2155 Matrix

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

POJ Monthly,Lou Tiancheng

 

解题：二维线段树，第一次玩这鬼玩意，调试了N久。。。挫。。。。

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct subtree{
     int lt,rt;
     bool val;
 };
 struct node{
     int lt,rt;
     subtree sb[maxn<<];
 };
 node tree[maxn<<];
 int n,m,ans;
 void sub(int lt,int rt,int v,subtree *sb){
     sb[v].lt = lt;
     sb[v].rt = rt;
     sb[v].val = false;
     if(lt == rt) return;
     int mid = (lt + rt)>>;
     sub(lt,mid,v<<,sb);
     sub(mid+,rt,v<<|,sb);
 }
 void build(int lt,int rt,int v){
     tree[v].lt = lt;
     tree[v].rt = rt;
     sub(,n,,tree[v].sb);
     if(lt == rt) return;
     int mid = (lt + rt)>>;
     build(lt,mid,v<<);
     build(mid+,rt,v<<|);
 }
 void subupdate(int lt,int rt,int v,subtree *sb){
     if(sb[v].lt >= lt && sb[v].rt <= rt){
         sb[v].val ^= ;
         return;
     }
     if(lt <= tree[v<<].rt) subupdate(lt,rt,v<<,sb);
     if(rt >= tree[v<<|].lt) subupdate(lt,rt,v<<|,sb);
 }
 void update(int lt,int rt,int y1,int y2,int v){
     if(tree[v].lt >= lt && tree[v].rt <= rt){
         subupdate(y1,y2,,tree[v].sb);
         return;
     }
     if(lt <= tree[v<<].rt) update(lt,rt,y1,y2,v<<);
     if(rt >= tree[v<<|].lt) update(lt,rt,y1,y2,v<<|);

 }
 void subquery(int y,int v,subtree *sb){
     ans ^= sb[v].val;
     if(sb[v].lt == sb[v].rt) return;
     if(y <= sb[v<<].rt) subquery(y,v<<,sb);
     else subquery(y,v<<|,sb);
 }
 void query(int x,int y,int v){
     subquery(y,,tree[v].sb);
     if(tree[v].lt == tree[v].rt) return;
     if(x <= tree[v<<].rt) query(x,y,v<<);
     else query(x,y,v<<|);
 }
 int main(){
     int t,x1,y1,x2,y2;
     char s[];
     scanf("%d",&t);
     while(t--){
         scanf("%d %d",&n,&m);
         build(,n,);
         for(int i = ; i < m; ++i){
             scanf("%s",s);
             if(s[] == 'Q'){
                 ans = ;
                 scanf("%d %d",&x1,&y1);
                 query(x1,y1,);
                 printf("%d\n",ans);
             }else if(s[] == 'C'){
                 scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
                 update(x1,x2,y1,y2,);
             }
         }
         puts("");
     }
     return ;
 }