poj 2007 Scrambled Polygon (凸包 )

题意：

给出若干个点，求出它们的凸包，并且按原点为第一点的逆时针方向输出。

 

输出为：

(0,0)
(-30,-40) 
(-30,-50) 
(-10,-60) 
(50,-60) 
(70,-50) 
(90,-20) 
(90,10) 
(80,20) 
(60,30)

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<set>
 7 #include<map>
 8 #include<queue>
 9 #include<vector>
10 #include<string>
11 #define Min(a,b) a<b?a:b
12 #define Max(a,b) a>b?a:b
13 #define CL(a,num) memset(a,num,sizeof(a));
14 #define maxn  60
15 #define eps  1e-12
16 #define inf 100000000
17 #define mx 1<<60
18 #define ll   __int64
19 using namespace std;
20 struct point
21 {
22     double x,y;
23 }p[maxn];
24 int res[maxn];
25 int dblcmp(double x)
26 {
27     if(fabs(x) < eps) return 0;
28     if(x < 0) return -1;
29     else return 1;
30 }
31 double det(double x1,double y1,double x2,double y2)
32 {
33     return x1*y2 - x2*y1 ;
34 }
35 double cross(point a, point b, point c)
36 {
37     return det(b.x - a.x, b.y - a.y, c.x - a.x, c.y - a.y);
38 }
39 int cmp(point a,point b)
40 {
41     if(a.y != b.y) return a.y < b.y;
42     else return a.x < b.x;
43 }
44 int  top ,n;
45 void graham()
46 {
47     int  i,len;
48     top = 0;
49     sort(p,p+n,cmp);
50     
51     if(n == 0) return ;res[top++] = 0;
52     if(n == 1) return ;res[top++] = 1;
53    
54     for(i = 2 ;i<n;i++)//求右链
55     {
56         while(top&&dblcmp(cross(p[res[top - 1]],p[res[top - 2]],p[i])) > 0) top--;
57 
58         res[top++] = i;
59     }
60      
61     len =  top ;
62     
63     for(i = n - 2;i >= 0;i--)//求左链
64     {
65          while(top != len && dblcmp(cross(p[res[top - 1]],p[res[top - 2]],p[i])) > 0)top--;
66          res[top++] = i;
67 
68     }
69     top--;//第一个点 入栈两次 所以 减 1；
70 
71 }
72 int main()
73 {
74 
75       n = 0;
76       int  i,s;
77     //freopen("data.txt","r",stdin);
78     while(scanf("%lf%lf",&p[n].x,&p[n].y)!=EOF)n++;
79     graham() ;
80 
81     for(s = 0 ; s < top;s++)
82     {
83         if(p[res[s]].x == 0&&p[res[s]].y == 0) {break;}
84     }
85     for( i = s ; i < top; i++)
86     {
87         printf("(%.0lf,%.0lf)\n",p[res[i]].x,p[res[i]].y);
88     }
89     for(i =0 ;  i< s;i++)
90     {
91          printf("(%.0lf,%.0lf)\n",p[res[i]].x,p[res[i]].y);
92     }
93 
94 }