poj 3277 Mountains

最新推荐文章于 2020-03-06 15:47:33 发布

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最新推荐文章于 2020-03-06 15:47:33 发布

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原文链接：http://www.cnblogs.com/fanminghui/p/3574263.html

本文介绍了一个计算几何问题的解决方法，通过使用C++实现特定算法来计算一条折线路径的总长度。该路径由一系列坐标点定义，并考虑了直线段之间的交点计算，确保了路径的准确性和完整性。

http://poj.org/problem?id=3227

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 #define maxn 2000
 6 using namespace std;
 7 
 8 int n,h;
 9 const double eps=1e-8;
10 int cmp(double x)
11 {
12     if(fabs(x)<eps) return 0;
13     if(x>0) return 1;
14     return -1;
15 }
16 struct point
17 {
18     double x,y;
19     point() {}
20     point(double a,double b):x(a),y(b) {}
21 } p[maxn];
22 
23 double sqr(double x)
24 {
25     return x*x;
26 }
27 
28 double dis(point a,point b)
29 {
30     return (sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)));
31 }
32 
33 double det(point a,point b,point c,point d)
34 {
35     return (b.x-a.x)*(d.y-c.y)-(b.y-a.y)*(d.x-c.x);
36 }
37 
38 bool segem(point a,point b,point c,point d,point &e)
39 {
40     double s1=det(a,b,a,c);
41     double s2=det(a,b,a,d);
42     e.x=(c.x*s2-d.x*s1)/(s2-s1);
43     e.y=(c.y*s2-d.y*s1)/(s2-s1);
44     return true;
45 }
46 
47 double kk(point a,point b)
48 {
49     return (b.y-a.y)/(b.x-a.x);
50 }
51 int main()
52 {
53     while(scanf("%d%d",&n,&h)!=EOF)
54     {
55         if(n==0&&h==0) break;
56         for(int i=0; i<n; i++)
57         {
58             scanf("%lf%lf",&p[i].x,&p[i].y);
59         }
60         point res;
61         res.x=0;
62         res.y=h;
63         double sum=0;
64         sum+=dis(p[0],p[1]);
65         point key;
66         key=p[1];
67         for(int i=2; i<n; i++)
68         {
69             double k3=kk(res,p[i]);
70             double k1=kk(p[i],p[i-1]);
71             double k2=kk(res,key);
72             if(k3-k2>eps)
73             {
74                 if(k1-k2>=0)
75                 {
76                     point e;
77                     segem(res,key,p[i-1],p[i],e);
78                     sum+=dis(e,p[i]);
79                     key=p[i];
80                 }
81             }
82         }
83         printf("%.2lf\n",sum);
84     }
85     return 0;
86 }