POJ 2155 Matrix （二维线段树）

Matrix

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 17226		Accepted: 6461

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

这题二维线段树也可以做。

二维线段树需要  给一个矩形加一个值

查询单个的值。

加值的时候直接加一个块。

查询的时候把这个点以及和这个点相关的都累加起来。

数据结构写法多样啊，重在理解

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2014/5/23 23:08:19
 File Name     :E:\2014ACM\专题学习\数据结构\二维线段树\POJ2155.cpp
  ************************************************ */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MAXN = ;
 struct Nodey
 {
     int l,r;
     int val;
 };
 int n;
 int locx[MAXN],locy[MAXN];
 struct Nodex
 {
     int l,r;
     Nodey sty[MAXN*];
     void build(int i,int _l,int _r)
     {
         sty[i].l = _l;
         sty[i].r = _r;
         sty[i].val = ;
         if(_l == _r)
         {
             locy[_l] = i;
             return;
         }
         int mid = (_l + _r)>>;
         build(i<<,_l,mid);
         build((i<<)|,mid+,_r);
     }
     void add(int i,int _l,int _r,int val)
     {
         if(sty[i].l == _l && sty[i].r == _r)
         {
             sty[i].val += val;
             return;
         }
         int mid = (sty[i].l + sty[i].r)>>;
         if(_r <= mid)add(i<<,_l,_r,val);
         else if(_l > mid)add((i<<)|,_l,_r,val);
         else
         {
             add(i<<,_l,mid,val);
             add((i<<)|,mid+,_r,val);
         }
     }
 }stx[MAXN*];
 void build(int i,int l,int r)
 {
     stx[i].l = l;
     stx[i].r = r;
     stx[i].build(,,n);
     if(l == r)
     {
         locx[l] = i;
         return;
     }
     int mid = (l+r)>>;
     build(i<<,l,mid);
     build((i<<)|,mid+,r);
 }
 void add(int i,int x1,int x2,int y1,int y2,int val)
 {
     if(stx[i].l == x1 && stx[i].r == x2)
     {
         stx[i].add(,y1,y2,val);
         return;
     }
     int mid = (stx[i].l + stx[i].r)/;
     if(x2 <= mid)add(i<<,x1,x2,y1,y2,val);
     else if(x1 > mid)add((i<<)|,x1,x2,y1,y2,val);
     else
     {
         add(i<<,x1,mid,y1,y2,val);
         add((i<<)|,mid+,x2,y1,y2,val);
     }
 }
 int sum(int x,int y)
 {
     int ret = ;
     for(int i = locx[x];i;i >>= )
         for(int j = locy[y];j;j >>= )
             ret += stx[i].sty[j].val;
     return ret;
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int q;
         scanf("%d%d",&n,&q);
         build(,,n);
         char op[];
         int x1,x2,y1,y2;
         while(q--)
         {
             scanf("%s",op);
             if(op[] == 'C')
             {
                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                 add(,x1,x2,y1,y2,);
             }
             else
             {
                 scanf("%d%d",&x1,&y1);
                 if(sum(x1,y1)% == )printf("0\n");
                 else printf("1\n");
             }
         }
         if(T)printf("\n");
     }
     return ;
 }