求平行四边形的第四个点的坐标

最新推荐文章于 2020-12-31 09:04:13 发布

转载最新推荐文章于 2020-12-31 09:04:13 发布 · 1.7k 阅读

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原文链接：http://www.cnblogs.com/ccut-ry/p/8995055.html

文章标签：

#c/c++

本文介绍了一个计算几何问题的解决方案，即通过已知平行四边形相邻两边的两个端点坐标来求解第四点坐标的方法。利用向量运算原理，提供了具体的C++代码实现。

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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 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145

Sample Output

1.000 0.000
-2.500 -2.500
0.151 -0.398

题意 ：给你平行四边形的相邻的两条边，让你求第四个点的坐标
思路分析：一个比较基础的计算几何题，只要根据向量的矢量和相等即可求解
　　　　相等的点的位置有 4 种情况，分别是 1 3、 1 4、 2 3、 2 4分别求即可
代码示例 ：

struct point
{
    double x, y;
    point(double _x=0, double _y=0):x(_x), y(_y){}
     
    // 点-点=向量
    point operator-(const point &v){
        return point(x-v.x, y-v.y);
    }
 
};
 
int dcmp(double x){
    if (fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
bool operator == (const point &a, const point &b){
    return (dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0);
}
 
typedef point Vector; // Vector表示向量

point p[10];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    while(~scanf("%lf%lf", &p[1].x, &p[1].y)){
        for(int i = 2; i <= 4; i++) {
            scanf("%lf%lf", &p[i].x, &p[i].y);
        }
        
        if (p[2] == p[3]) {
            Vector v = p[4] - p[3];
            printf("%.3f %.3f\n", v.x+p[1].x, v.y+p[1].y);
        }
        if (p[2] == p[4]){
            Vector v = p[3] - p[4];
            printf("%.3f %.3f\n", v.x+p[1].x, v.y+p[1].y);
        }
        if (p[1] == p[3]){
            Vector v = p[4]-p[3];
            printf("%.3f %.3f\n", v.x+p[2].x, v.y+p[2].y);
        }
        if (p[1] == p[4]){
            Vector v = p[3] - p[4];
            printf("%.3f %.3f\n", v.x+p[2].x, v.y+p[2].y);
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/ccut-ry/p/8995055.html