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hdu2215（最小覆盖圆）
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 ?
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.
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 else101 MiniDiscWithPoint(a[i],i-1);102 }103 printf("%.2lf\n",r+0.5);104 }105 return 0;106 }
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