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hdu2215（最小覆盖圆）

2024-07-16 15:14:33   221人阅读

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