Maple trees
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1578 Accepted Submission(s):
488
Problem Description
There are a lot of trees in HDU. Kiki want to surround
all the trees with the minimal required length of the rope . As follow, 
To make
this problem more simple, consider all the trees are circles in a plate. The
diameter of all the trees are the same (the diameter of a tree is 1 unit). Kiki
can calculate the minimal length of the rope , because it‘s so easy for this
smart girl.
But we don‘t have a rope to surround the trees. Instead, we only
have some circle rings of different radius. Now I want to know the minimal
required radius of the circle ring. And I don‘t want to ask her this problem,
because she is busy preparing for the examination.
As a smart ACMer, can you
help me ?
Input
The input contains one or more data sets. At first line
of each input data set is number of trees in this data set n （1 <= n <=
100）, it is followed by n coordinates of the trees. Each coordinate is a pair of
integers, and each integer is in [-1000, 1000], it means the position of a
tree’s center. Each pair is separated by blank.
Zero at line for number of
trees terminates the input for your program.
Output
Minimal required radius of the circle ring I have to
choose. The precision should be 10^-2.
Sample Input
2
1 0
-1 0
0
Sample Output
1.50
题意：用一个最小的圆把所有的点都圈在里边，点可以在圆上，每个点的半径为0.50
1 #include<stdio.h>
2 #include<math.h>
3 #define PI acos(-1.0)
4 struct TPoint
5 {
6 double x,y;
7 }a[1005],d;
8 double r;
9 double distance(TPoint p1, TPoint p2)
10 {
11 return (sqrt((p1.x-p2.x)*(p1.x -p2.x)+(p1.y-p2.y)*(p1.y-p2.y)));
12 }
13 double multiply(TPoint p1, TPoint p2, TPoint p0)
14 {
15 return ((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y));
16 }
17 void MiniDiscWith2Point(TPoint p,TPoint q,int n)
18 {
19 d.x=(p.x+q.x)/2.0;
20 d.y=(p.y+q.y)/2.0;
21 r=distance(p,q)/2;
22 int k;
23 double c1,c2,t1,t2,t3;
24 for(k=1; k<=n; k++)
25 {
26 if(distance(d,a[k])<=r)
27 continue;
28 if(multiply(p,q,a[k])!=0.0)
29 {
30 c1=(p.x*p.x+p.y*p.y-q.x*q.x-q.y*q.y)/2.0;
31 c2=(p.x*p.x+p.y*p.y-a[k].x*a[k].x-a[k].y*a[k].y)/2.0;
32
33 d.x=(c1*(p.y-a[k].y)-c2*(p.y-q.y))/((p.x-q.x)*(p.y-a[k].y)-(p.x-a[k].x)*(p.y-q.y));
34 d.y=(c1*(p.x-a[k].x)-c2*(p.x-q.x))/((p.y-q.y)*(p.x-a[k].x)-(p.y-a[k].y)*(p.x-q.x));
35 r=distance(d,a[k]);
36 }
37 else
38 {
39 t1=distance(p,q);
40 t2=distance(q,a[k]);
41 t3=distance(p,a[k]);
42 if(t1>=t2&&t1>=t3)
43 {
44 d.x=(p.x+q.x)/2.0;
45 d.y=(p.y+q.y)/2.0;
46 r=distance(p,q)/2.0;
47 }
48 else if(t2>=t1&&t2>=t3)
49 {
50 d.x=(a[k].x+q.x)/2.0;
51 d.y=(a[k].y+q.y)/2.0;
52 r=distance(a[k],q)/2.0;
53 }
54 else
55 {
56 d.x=(a[k].x+p.x)/2.0;
57 d.y=(a[k].y+p.y)/2.0;
58 r=distance(a[k],p)/2.0;
59 }
60 }
61 }
62 }
63
64 void MiniDiscWithPoint(TPoint pi,int n)
65 {
66 d.x=(pi.x+a[1].x)/2.0;
67 d.y=(pi.y+a[1].y)/2.0;
68 r=distance(pi,a[1])/2.0;
69 int j;
70 for(j=2; j<=n; j++)
71 {
72 if(distance(d,a[j])<=r)
73 continue;
74 else
75 {
76 MiniDiscWith2Point(pi,a[j],j-1);
77 }
78 }
79 }
80 int main()
81 {
82 int i,n;
83 while(scanf("%d",&n)&&n)
84 {
85 for(i=1; i<=n; i++)
86 scanf("%lf %lf",&a[i].x,&a[i].y);
87 if(n==1)
88 {
89 printf("0.50\n");
90 continue;
91 }
92
93 r=distance(a[1],a[2])/2.0;
94 d.x=(a[1].x+a[2].x)/2.0;
95 d.y=(a[1].y+a[2].y)/2.0;
96 for(i=3; i<=n; i++)
97 {
98 if(distance(d,a[i])<=r)
99 continue;
100 else
101 MiniDiscWithPoint(a[i],i-1);
102 }
103 printf("%.2lf\n",r+0.5);
104 }
105 return 0;
106 }
hdu2215（最小覆盖圆）