hdu1392凸包裸题

 1 //极角排序
 2 #include <bits/stdc++.h>
 3 #define sqr(x) ((x)*(x))
 4 using namespace std;
 5 int n,st[200001],top;
 6 struct POI
 7 {
 8     int x,y;
 9     POI()
10     {
11         x=y=0;
12     }
13     POI(int _x,int _y)
14     {
15         x=_x;y=_y;
16     }
17 } p[200001];
18 int cross(POI x,POI y)
19 {
20     return x.x*y.y-x.y*y.x;
21 }
22 double dis(POI x,POI y)
23 {
24     return sqrt(sqr(x.x-y.x)+sqr(x.y-y.y));
25 }
26 POI operator-(POI x,POI y)
27 {
28     return POI(x.x-y.x,x.y-y.y);
29 }
30 bool com(POI x,POI y)
31 {
32     return(cross(x-p[1],y-p[1])>0 || cross(x-p[1],y-p[1])==0 && dis(x,p[1])<dis(y,p[1]));
33 }
34 int main()
35 {
36     for(scanf("%d",&n);n;scanf("%d",&n))
37     {
38         for(int i=1;i<=n;i++)
39             scanf("%d%d",&p[i].x,&p[i].y);
40         if(n==2)//wtf 
41         {
42             printf("%.2f\n",dis(p[1],p[2]));
43             continue;
44         }
45         int id=1;
46         for(int i=2;i<=n;i++)
47             if(p[i].x<p[id].x || p[i].x==p[id].x && p[i].y<p[id].y) id=i;
48         swap(p[1],p[id]);
49         sort(p+2,p+n+1,com);
50         st[top=1]=st[0]=1;
51         for(int i=2;i<=n;i++)
52         {
53             while(top && cross(p[st[top]]-p[st[top-1]],p[i]-p[st[top]])<0) --top;
54             st[++top]=i;
55         }
56         double len=0;
57 //        puts("");
58 //        for(int i=1;i<=top;i++)
59 //            printf("%d %d\n",p[st[i]].x,p[st[i]].y);
60         st[++top]=1;
61         for(int i=1;i<top;i++)
62             len+=dis(p[st[i]],p[st[i+1]]);
63         printf("%.2f\n",len);
64     }
65     return 0;
66 }

 

posted @ 2017-07-09 16:27  汪立超  阅读(101)  评论(0编辑  收藏