nyoj-1132-promise me a medal(求线段交点)

题目链接

 /*
     Name:nyoj-1132-promise me a medal
     Copyright:
     Author:
     Date: 2018/4/26 20:26:22
     Description:
     向量之间的运算，不熟悉就绕蒙逼了
     用 ACM国际大学生程序设计竞赛 算法与实现的模板
     有个坑：    如果第二条线段的起点是第一条线段的终点，那么不需要计算  

     1
     1 2 1 3 1 3 1 4
     输出
     yes 1.0 3.0 

 */
 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 const double pi = acos(-1.0);
 inline double sqr(double x) {
     return x * x;
 }
 const double eps = 1e-;
 int cmp(double x) {
     if (fabs(x) < eps) return ;
     if (x > )return ;
     return -;
 }
 //point
 struct point {
     double x, y;
     point (){}
     point (double a, double b):x(a), y(b) {    }
     void input() {
         scanf("%lf %lf", &x, &y);
     }
     friend point operator - (const point &a, const point &b) {
         return point(a.x-b.x, a.y-b.y);
     }
     friend point operator * (const double &a, const point &b) {
         return point (a * b.x, a*b.y);
     }
     friend point operator / (const point &a, const double &b) {
         return point (a.x / b, a.y /b);
     }
     friend bool operator == (const point &a, const point &b) {
         return (cmp(a.x - b.x) ==  && cmp(a.y - b.y) == );
     }
 };
 double det(const point &a, const point &b) {
     return a.x * b.y - a.y * b.x;
 }
 //line
 struct line {
     point a, b;
     line() {    }
     line(point x, point y):a(x), b(y){    }
 };
 line point_make_line(const point a, const point b) {
     return line(a, b);
 }
 bool parallel(line a, line b)  {
     return !cmp(det(a.a - a.b, b.a - b.b));
 }
 bool line_make_point(line a, line b, point &res) {
     if(parallel(a,b)) return false;
     double s1 = det(a.a-b.a, b.b - b.a);
     double s2 = det(a.b-b.a, b.b - b.a);
     res = (s1 * a.b - s2 * a.a)/(s1-s2);
     return true;
 }
 int main()
 {
     int t;
     cin>>t;
     while (t--) {
         point a, b ,c ,d;
         cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>d.y;
         line ab(a, b), cd(c, d);
         if (b == c) {//如果第二条线段的起点是第一条线段的终点，那么不需要计算
             cout<<"yes ";
             printf("%.1f %.1f\n",b.x, b.y) ;
             continue;
         }
         if (parallel(ab, cd)) cout<<"no\n";
         else {
             cout<<"yes ";
             point ans;
             line_make_point(ab, cd, ans);
             printf("%.1f %.1f\n",ans.x, ans.y) ;
         }
     }
     return ;
 }