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  • 求平行四边形的第四个点的坐标

    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
    ", v.x+p[1].x, v.y+p[1].y);
            }
            if (p[2] == p[4]){
                Vector v = p[3] - p[4];
                printf("%.3f %.3f
    ", v.x+p[1].x, v.y+p[1].y);
            }
            if (p[1] == p[3]){
                Vector v = p[4]-p[3];
                printf("%.3f %.3f
    ", v.x+p[2].x, v.y+p[2].y);
            }
            if (p[1] == p[4]){
                Vector v = p[3] - p[4];
                printf("%.3f %.3f
    ", v.x+p[2].x, v.y+p[2].y);
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/ccut-ry/p/8995055.html
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