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

Sample Output

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 ;

 }

巴特西