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poj 2155:Matrix(二维线段树,矩阵取反,好题)

Matrix
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 17880   Accepted: 6709

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
 
  二维线段树,矩阵取反,好题
 
  题意
  
  思路
  矩阵节点的值为是否取反,0为不取反,1为取反,暂称为取反值。
  取反操作的时候先找到这个矩阵代表的节点,然后将这个节点的值+1再模2,即取反。
  查询的时候,将(x,y)这个坐标经过的所有矩阵的取反值加起来,每次%2,最后那个值就为这个坐标最后的值。
  为什么%2,因为不是1就是0,矩阵记录了取反值,找这个坐标的过程中,经过的矩阵如果取反值为1,则结果变为0,在经过一个取反值为1的矩阵,结果又变为1…… 直到加到要找的坐标的取反值,这个结果记录的值就是这个坐标的取反值。这个时候输出结果。
 
  代码
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 
 6 #define MAXN 1100
 7 
 8 int tree[MAXN*3][MAXN*3],s;
 9 
10 void Negate_y(int d,int dy,int L,int R,int y1,int y2)    //取反操作
11 {    
12     if(L==y1 && R==y2){    //将这个矩阵的所有元素记录为取反
13         tree[d][dy] = (tree[d][dy]+1) % 2;
14         return ;
15     }
16 
17     int mid = (L+R)>>1;
18     if(mid>=y2)
19         Negate_y(d,dy<<1,L,mid,y1,y2);
20     else if(mid<y1)
21         Negate_y(d,dy<<1|1,mid+1,R,y1,y2);
22     else{
23         Negate_y(d,dy<<1,L,mid,y1,mid);
24         Negate_y(d,dy<<1|1,mid+1,R,mid+1,y2);
25     }
26 }
27 
28 void Negate_x(int d,int L,int R,int x1,int y1,int x2,int y2)        //取反操作
29 {
30     if(L==x1 && R==x2){    //找到行块
31         Negate_y(d,1,1,s,y1,y2);    
32         return ;
33     }
34 
35     int mid = (L+R)>>1;
36     if(mid>=x2)
37         Negate_x(d<<1,L,mid,x1,y1,x2,y2);
38     else if(mid<x1)
39         Negate_x(d<<1|1,mid+1,R,x1,y1,x2,y2);
40     else{
41         Negate_x(d<<1,L,mid,x1,y1,mid,y2);
42         Negate_x(d<<1|1,mid+1,R,mid+1,y1,x2,y2);
43     }
44 }
45 
46 
47 int Query_y(int d,int dy,int L,int R,int r)    //查询
48 {
49     if(L==R)    //找到要找的坐标,输出这个坐标对应的值
50         return tree[d][dy];
51 
52     //没找到
53     int mid = (L+R)>>1;
54     if(mid >= r)
55         return (Query_y(d,dy<<1,L,mid,r)+tree[d][dy]) % 2;
56     else
57         return (Query_y(d,dy<<1|1,mid+1,R,r)+tree[d][dy]) % 2;
58 }
59 
60 int Query_x(int d,int L,int R,int l,int r)    //查询
61 {
62     if(L==R){    //找到要找的行块,继续查找列块
63         return Query_y(d,1,1,s,r);
64     }
65 
66     //没找到
67     int mid = (L+R)>>1;
68     if(mid >= l)
69         return (Query_x(d<<1,L,mid,l,r) + Query_y(d,1,1,s,r)) % 2;
70     else
71         return (Query_x(d<<1|1,mid+1,R,l,r) + Query_y(d,1,1,s,r)) % 2;
72 }
73 
74 int main()
75 {
76     int X,T,x,y,x1,y1,x2,y2;
77     scanf("%d",&X);
78     while(X--){
79         memset(tree,0,sizeof(tree));
80         scanf("%d%d",&s,&T);
81         while(T--){
82             char c[5];
83             scanf("%s",c);
84             if(c[0]=='C'){
85                 scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
86                 Negate_x(1,1,s,x1,y1,x2,y2);
87             }
88             else if(c[0]=='Q'){
89                 scanf("%d%d",&x,&y);
90                 printf("%d\n",Query_x(1,1,s,x,y));
91             }
92         }
93         if(X!=0)
94             printf("\n");
95     }
96 
97     return 0;
98 }

 

Freecode : www.cnblogs.com/yym2013

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