题目链接
1 /*
2 Name:nyoj-1132-promise me a medal
3 Copyright:
4 Author:
5 Date: 2018/4/26 20:26:22
6 Description:
7 向量之间的运算，不熟悉就绕蒙逼了
8 用 ACM国际大学生程序设计竞赛 算法与实现的模板
9 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
10
11 1
12 1 2 1 3 1 3 1 4
13 输出
14 yes 1.0 3.0
15
16 */
17 #include <iostream>
18 #include <cstdio>
19 #include <cmath>
20 #include <algorithm>
21 using namespace std;
22 const double pi = acos(-1.0);
23 inline double sqr(double x) {
24 return x * x;
25 }
26 const double eps = 1e-8;
27 int cmp(double x) {
28 if (fabs(x) < eps) return 0;
29 if (x >0 )return 1;
30 return -1;
31 }
32 //point
33 struct point {
34 double x, y;
35 point (){}
36 point (double a, double b):x(a), y(b) { }
37 void input() {
38 scanf("%lf %lf", &x, &y);
39 }
40 friend point operator - (const point &a, const point &b) {
41 return point(a.x-b.x, a.y-b.y);
42 }
43 friend point operator * (const double &a, const point &b) {
44 return point (a * b.x, a*b.y);
45 }
46 friend point operator / (const point &a, const double &b) {
47 return point (a.x / b, a.y /b);
48 }
49 friend bool operator == (const point &a, const point &b) {
50 return (cmp(a.x - b.x) == 0 && cmp(a.y - b.y) == 0);
51 }
52 };
53 double det(const point &a, const point &b) {
54 return a.x * b.y - a.y * b.x;
55 }
56 //line
57 struct line {
58 point a, b;
59 line() { }
60 line(point x, point y):a(x), b(y){ }
61 };
62 line point_make_line(const point a, const point b) {
63 return line(a, b);
64 }
65 bool parallel(line a, line b) {
66 return !cmp(det(a.a - a.b, b.a - b.b));
67 }
68 bool line_make_point(line a, line b, point &res) {
69 if(parallel(a,b)) return false;
70 double s1 = det(a.a-b.a, b.b - b.a);
71 double s2 = det(a.b-b.a, b.b - b.a);
72 res = (s1 * a.b - s2 * a.a)/(s1-s2);
73 return true;
74 }
75 int main()
76 {
77 int t;
78 cin>>t;
79 while (t--) {
80 point a, b ,c ,d;
81 cin>>a.x>>a.y>>b.x>>b.y>>c.x>>c.y>>d.x>>d.y;
82 line ab(a, b), cd(c, d);
83 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
84 cout<<"yes ";
85 printf("%.1f %.1f\n",b.x, b.y) ;
86 continue;
87 }
88 if (parallel(ab, cd)) cout<<"no\n";
89 else {
90 cout<<"yes ";
91 point ans;
92 line_make_point(ab, cd, ans);
93 printf("%.1f %.1f\n",ans.x, ans.y) ;
94 }
95 }
96 return 0;
97 }
