POJ-2624 4th Point（计算几何）

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4th Point

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4943 Accepted: 1735

Description

Given are the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. Find the (x,y) coordinates of the fourth point.

Input

Each line of input contains eight floating point numbers: the (x,y) coordinates of one of the endpoints of the first side followed by the (x,y) coordinates of the other endpoint of the first side, followed by the (x,y) coordinates of one of the endpoints of the second side followed by the (x,y) coordinates of the other endpoint of the second side. All coordinates are in meters, to the nearest mm. All coordinates are between -10000 and +10000.

Output

For each line of input, print the (x,y) coordinates of the fourth point of the parallelogram in meters, to the nearest mm, separated by a single space.

Sample Input

0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.0001.000 0.000 3.500 3.500 3.500 3.500 0.000 1.0001.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145

Sample Output

1.000 0.000-2.500 -2.5000.151 -0.398

我勒个fuck，这个貌似也是调了好久，要赶紧将状态提上来啊

题意：已知平行四边形的两条邻边，求第四个点的坐标

思路：相同的那个点为基础点（自己定义的），用其他两个点与其形成两个向量，加和即平行四边形对角线向量，再加基础点就ok了。。点+向量=点按照向量的方向延长向量的模的长度

就像这个，B+BD向量=D

#include <iostream>#include<cstdio>#include <cmath>#include <algorithm>using namespace std;const double EPS = 1e-6;struct Point{    double x,y;    Point(){};    Point(double a,double b){        x=a;y=b;    }};Point operator - (const Point a,const Point b){        return Point(a.x-b.x,a.y-b.y);}Point operator + (const Point a,const Point b){       return Point(a.x+b.x,a.y+b.y);}int sgn(double d){  return (d>+EPS)-(d<-EPS);}int iszero(Point a,Point b){    if (sgn(a.x-b.x)==0&&sgn(a.y-b.y)==0) return 1;    else return 0;}Point p,q,r,s,t;int main(){//p,r,t,q    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p.x,&p.y,&r.x,&r.y,&t.x,&t.y,&q.x,&q.y))    {        if(iszero(p,t))            s = p + r - p + q -t;        if(iszero(p,q))            s = p + r - p + t - q;        if(iszero(r,t))            s = r + p - r + q - t;        if(iszero(r,q))            s = r + p - r + t - q;        printf("%.3f %.3f\n",s.x,s.y);    }    return 0;}

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