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

 

2
0 0
0 2
3 0
-1 1
0 0
1 1

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

#include<iostream>

using namespace std;

typedef struct point
{
    long x;
    long y;
} POINT;

typedef struct vector
{
    long x;
    long y;
} VECTOR;

POINT f(POINT l,POINT r,POINT m)
{
    POINT mid,result;
    mid.x=l.x+r.x;
    mid.y=l.y+r.y;
    result.x=mid.x-m.x;
    result.y=mid.y-m.y;
    return result;
}

int main()
{
    int t;
    POINT a,b,c,d;
    VECTOR m,n,p;

    cin>>t;

    while(t--)
    {
        cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y;
        m.x=a.x-b.x;
        m.y=a.y-b.y;
        n.x=b.x-c.x;
        n.y=b.y-c.y;
        p.x=c.x-a.x;
        p.y=c.y-a.y;
        if(m.x*n.x+m.y*n.y==0)
            d=f(a,c,b);
        else if(n.x*p.x+n.y*p.y==0)
            d=f(a,b,c);
        else if(p.x*m.x+p.y*m.y==0)
            d=f(b,c,a);
        cout<<d.x<<" "<<d.y<<endl;
    }

    return 0;
}