An Easy Problem?!（细节题，要把所有情况考虑到）

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <algorithm> 
 4 using namespace std;
 5 
 6 #define N 105
 7 #define eps 1e-6//精度太小会导致出错，精度太大会增加计算量，自己权衡 
 8 struct point {
 9     double x, y;
10     point(){}
11     point(double x, double y):x(x), y(y){}
12     point operator + (const point o) const{
13         return point(x+o.x, y+o.y);
14     }
15     point operator - (const point o) const{
16         return point(x-o.x, y-o.y);
17     }
18 
19     double operator * (const point o) const{
20         return x*o.y - o.x*y;
21     }
22 
23     double operator ^ (const point o) const{//点乘
24         return x*o.x + y*o.y;
25     }
26 
27     point operator * (const double a) const{
28         return point(a*x, a*y);
29     }
30 
31     double len2()
32     {
33         return x*x + y*y;
34     }
35 }a, b, s, t;
36 
37 point Intersection(point a, point b, point c, point d)
38 {
39     double t = ((d - a)*(c - a))/((b - a)*(d - c));
40     return a + (b-a)*fabs(t);
41 }
42 
43 int main()
44 {
45     point a, b, c, d;
46     int T;
47     scanf("%d", &T);
48     while(T--)
49     {
50         scanf("%lf %lf %lf %lf", &a.x, &a.y, &b.x, &b.y);
51         scanf("%lf %lf %lf %lf", &c.x, &c.y, &d.x, &d.y);
52         
53         if(fabs(a.y - b.y) < eps || fabs(c.y - d.y) < eps)//有水平线 
54         {
55             puts("0.00");
56             continue;
57         }
58         
59         if(a.y < b.y) swap(a, b);
60         if(c.y < d.y) swap(c, d);
61         
62         if(fabs((b.y-a.y)*(d.x-c.x) - (d.y-c.y)*(b.x-a.x)) < eps) //两条线平行 
63         {
64             puts("0.00");
65             continue;
66         }
67         
68         if(((b-a)*(c-a))*((b-a)*(d-a)) > 0 || ((d-c)*(a-c))*((d-c)*(b-c)) > 0)//两条线段根本没有交点 
69         {
70             puts("0.00");
71             continue;
72         }
73         point p = Intersection(a, b, c, d); 
74         point up = point(0,10); 
75         if(((a-p)*(up)) * ((c-p)*(up)) > 0)//能收集雨水的部分在竖直线的同一侧
76         {
77             if((a-p)*(c-p) > 0 && c.x- a.x>= -eps)
78             {
79                 puts("0.00");
80                 continue;
81             }
82             if((a-p)*(c-p) < 0 && a.x- c.x>= -eps)
83             {
84                 puts("0.00");
85                 continue;
86             }
87         }
88         point t1, t2;
89         t1.y = t2.y = min(a.y, c.y);
90         t1.x = a.x + (b.x - a.x)*(t1.y - a.y)/(b.y-a.y);
91         t2.x = c.x + (d.x - c.x)*(t2.y - c.y)/(d.y-c.y);
92         double ans = fabs((t1.x - t2.x) * (t1.y-p.y)/2.0);
93         printf("%.2f\n", ans+eps); //控制精度， 
94     }
95     return 0;
96 }