hdu 3756 Dome of Circus(三分)

题意：求能将所有点覆盖的最小体积的圆锥。

体积变化为凹形，故可以三分半径r，再枚举各个点找最大的高h。

View Code

 1 /*

 2 Author:Zhaofa Fang

 3 Lang:C++

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 #include <sstream>

 8 #include <iostream>

 9 #include <cmath>

10 #include <cstring>

11 #include <algorithm>

12 #include <string>

13 #include <utility>

14 #include <vector>

15 #include <queue>

16 #include <stack>

17 #include <map>

18 #include <set>

19 using namespace std;

20 

21 typedef long long ll;

22 #define DEBUG(x) cout<< #x << ':' << x << endl

23 #define REP(i,n) for(int i=0;i < (n);i++)

24 #define REPD(i,n) for(int i=(n-1);i >= 0;i--)

25 #define FOR(i,s,t) for(int i = (s);i <= (t);i++)

26 #define FORD(i,s,t) for(int i = (s);i >= (t);i--)

27 #define PII pair<int,int>

28 #define PB push_back

29 #define MP make_pair

30 #define ft first

31 #define sd second

32 #define lowbit(x) (x&(-x))

33 #define INF (1<<30)

34 

35 

36 const double PI = 3.1415926;

37 int n;

38 struct point

39 {

40     double x,y,z;

41 }P[10005];

42 double volume(double r,double h)

43 {

44     return PI*r*r*h/3.0;

45 }

46 

47 double fun(double r,double &hx)

48 {

49     double MAX = 0;

50     REP(i,n)

51     {

52         double h = P[i].z*r/(r-sqrt(P[i].x*P[i].x+P[i].y*P[i].y));

53         MAX = max(MAX,h);

54     }

55     hx = MAX;

56     return volume(r,MAX);

57 }

58 int main()

59 {

60     //freopen("in","r",stdin);

61     //freopen("out","w",stdout);

62     int T;

63     scanf("%d",&T);

64     while(T--)

65     {

66         scanf("%d",&n);

67         double M=0;

68         REP(i,n)

69         {

70             scanf("%lf%lf%lf",&P[i].x,&P[i].y,&P[i].z);

71             M = max(M,sqrt(P[i].x*P[i].x+P[i].y*P[i].y));

72         }

73         double l=M+1e-6,r=100000.0,hx;

74         while(fabs(l-r) > 1e-6)

75         {

76             double lm = l + (r-l)/3.0;

77             double rm = r - (r-l)/3.0;

78             if(fun(lm,hx)<fun(rm,hx))r = rm;

79             else l = lm;

80         }

81         printf("%.3f %.3f\n",hx,r);

82     }

83     return 0;

84 }