BOJ 1562 Rectangle

Rectangle

Accept:11	Submit:14
Time Limit:1000MS	Memory Limit:65536KB

Rectangle

Description

Given three vertices of a rectangle on a 2D plane, you're required to determine the coordinate of the fourth vertex.

Input Format

An integer T(T≤100)
will exist in the first line of input, indicating the number of test cases.

Each test case consists of three lines, each with a pair of integers (x,y)
, which denotes a coordinate of a vertex.

All the coordinates will be set in range [−10000,10000]
. Note that the given vertices will exactly form a right angle.

Output Format

For each test case, output the coordinate of the required vertex.

Sample Input

Sample Output

3 2
0 2

给一个矩形的三个顶点，求第四个
我的做法是先用向量乘积为0的方法判断出给出的三点中的直角顶点，再由直角顶点和对角线中点确定出第四个顶点

 1 #include<iostream>
 2 
 3 using namespace std;
 4 
 5 typedef struct point
 6 {
 7     long x;
 8     long y;
 9 } POINT;
10 
11 typedef struct vector
12 {
13     long x;
14     long y;
15 } VECTOR;
16 
17 POINT f(POINT l,POINT r,POINT m)
18 {
19     POINT mid,result;
20     mid.x=l.x+r.x;
21     mid.y=l.y+r.y;
22     result.x=mid.x-m.x;
23     result.y=mid.y-m.y;
24     return result;
25 }
26 
27 int main()
28 {
29     int t;
30     POINT a,b,c,d;
31     VECTOR m,n,p;
32 
33     cin>>t;
34 
35     while(t--)
36     {
37         cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y;
38         m.x=a.x-b.x;
39         m.y=a.y-b.y;
40         n.x=b.x-c.x;
41         n.y=b.y-c.y;
42         p.x=c.x-a.x;
43         p.y=c.y-a.y;
44         if(m.x*n.x+m.y*n.y==0)
45             d=f(a,c,b);
46         else if(n.x*p.x+n.y*p.y==0)
47             d=f(a,b,c);
48         else if(p.x*m.x+p.y*m.y==0)
49             d=f(b,c,a);
50         cout<<d.x<<" "<<d.y<<endl;
51     }
52 
53     return 0;
54 }

[C++]