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

Sample Output

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

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2014/5/23 23:08:19
  4 File Name     :E:\2014ACM\专题学习\数据结构\二维线段树\POJ2155.cpp
  5  ************************************************ */
  6 
  7 #include <stdio.h>
  8 #include <string.h>
  9 #include <iostream>
 10 #include <algorithm>
 11 #include <vector>
 12 #include <queue>
 13 #include <set>
 14 #include <map>
 15 #include <string>
 16 #include <math.h>
 17 #include <stdlib.h>
 18 #include <time.h>
 19 using namespace std;
 20 const int MAXN = 1010;
 21 struct Nodey
 22 {
 23     int l,r;
 24     int val;
 25 };
 26 int n;
 27 int locx[MAXN],locy[MAXN];
 28 struct Nodex
 29 {
 30     int l,r;
 31     Nodey sty[MAXN*3];
 32     void build(int i,int _l,int _r)
 33     {
 34         sty[i].l = _l;
 35         sty[i].r = _r;
 36         sty[i].val = 0;
 37         if(_l == _r)
 38         {
 39             locy[_l] = i;
 40             return;
 41         }
 42         int mid = (_l + _r)>>1;
 43         build(i<<1,_l,mid);
 44         build((i<<1)|1,mid+1,_r);
 45     }
 46     void add(int i,int _l,int _r,int val)
 47     {
 48         if(sty[i].l == _l && sty[i].r == _r)
 49         {
 50             sty[i].val += val;
 51             return;
 52         }
 53         int mid = (sty[i].l + sty[i].r)>>1;
 54         if(_r <= mid)add(i<<1,_l,_r,val);
 55         else if(_l > mid)add((i<<1)|1,_l,_r,val);
 56         else
 57         {
 58             add(i<<1,_l,mid,val);
 59             add((i<<1)|1,mid+1,_r,val);
 60         }
 61     }
 62 }stx[MAXN*3];
 63 void build(int i,int l,int r)
 64 {
 65     stx[i].l = l;
 66     stx[i].r = r;
 67     stx[i].build(1,1,n);
 68     if(l == r)
 69     {
 70         locx[l] = i;
 71         return;
 72     }
 73     int mid = (l+r)>>1;
 74     build(i<<1,l,mid);
 75     build((i<<1)|1,mid+1,r);
 76 }
 77 void add(int i,int x1,int x2,int y1,int y2,int val)
 78 {
 79     if(stx[i].l == x1 && stx[i].r == x2)
 80     {
 81         stx[i].add(1,y1,y2,val);
 82         return;
 83     }
 84     int mid = (stx[i].l + stx[i].r)/2;
 85     if(x2 <= mid)add(i<<1,x1,x2,y1,y2,val);
 86     else if(x1 > mid)add((i<<1)|1,x1,x2,y1,y2,val);
 87     else 
 88     {
 89         add(i<<1,x1,mid,y1,y2,val);
 90         add((i<<1)|1,mid+1,x2,y1,y2,val);
 91     }
 92 }
 93 int sum(int x,int y)
 94 {
 95     int ret = 0;
 96     for(int i = locx[x];i;i >>= 1)
 97         for(int j = locy[y];j;j >>= 1)
 98             ret += stx[i].sty[j].val;
 99     return ret;
100 }
101 
102 int main()
103 {
104     //freopen("in.txt","r",stdin);
105     //freopen("out.txt","w",stdout);
106     int T;
107     scanf("%d",&T);
108     while(T--)
109     {
110         int q;
111         scanf("%d%d",&n,&q);
112         build(1,1,n);
113         char op[10];
114         int x1,x2,y1,y2;
115         while(q--)
116         {
117             scanf("%s",op);
118             if(op[0] == 'C')
119             {
120                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
121                 add(1,x1,x2,y1,y2,1);
122             }
123             else
124             {
125                 scanf("%d%d",&x1,&y1);
126                 if(sum(x1,y1)%2 == 0)printf("0\n");
127                 else printf("1\n");
128             }
129         }
130         if(T)printf("\n");
131     }
132     return 0;
133 }

posted on 2014-05-23 23:35 kuangbin 阅读(3721) 评论(0) 编辑收藏举报

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POJ 2155 Matrix （二维线段树）

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