hdu3264Open-air shopping malls(二分）

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枚举伞的圆心，最多只有20个，因为必须与某个现有的圆心重合。

然后再二分半径就可以了。

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 using namespace std;

11 #define N 25

12 #define LL long long

13 #define INF 0xfffffff

14 const double eps = 1e-8;

15 const double pi = acos(-1.0);

16 const double inf = ~0u>>2;

17 struct point

18 {

19     double x,y;

20     double r,s;

21     point(double x=0,double y=0):x(x),y(y) {}

22 } p[N];

23 int n;

24 typedef point pointt;

25 pointt operator -(point a,point b)

26 {

27     return point(a.x-b.x,a.y-b.y);

28 }

29 int dcmp(double x)

30 {

31     if(fabs(x)<eps) return 0;

32     return x<0?-1:1;

33 }

34 double circle_area(point a,point b)

35 {

36     double s,d,t,t1;

37     d=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

38     if(d>=a.r+b.r) s=0;

39     else if(d<=fabs(a.r-b.r)) s=min(acos(-1.0)*a.r*a.r,acos(-1.0)*b.r*b.r);

40     else

41     {

42         t=(a.r*a.r+d*d-b.r*b.r)/2.0/d;

43         t1=sqrt(a.r*a.r-t*t);

44         s=-d*t1+a.r*a.r*acos(t/a.r)+b.r*b.r*acos((d-t)/b.r);

45     }

46     return s;

47 }

48 int cal(double mid,point pp)

49 {

50     int i;

51     double sum;

52     pp.r = mid;

53     for(i = 1 ; i <= n ; i++)

54     {

55         sum = circle_area(pp,p[i]);

56         if(dcmp(sum-p[i].s/2)<0) return 0;

57     }

58     //cout<<mid<<" "<<pp.x<<" "<<pp.y<<" "<<pp.r<<endl;

59     return 1;

60 }

61 double solve(point pp)

62 {

63     double rig = 20001,lef = 0,mid;

64     while(rig-lef>eps)

65     {

66         mid = (rig+lef)/2;

67         if(cal(mid,pp))

68         rig = mid;

69         else lef = mid;

70     }

71     return rig;

72 }

73 int main()

74 {

75     int t,i;

76     cin>>t;

77     while(t--)

78     {

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

80         for(i = 1; i <= n ; i++)

81         {

82             scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].r);

83             p[i].s = pi*p[i].r*p[i].r;

84         }

85         double ans = INF;

86         for(i = 1; i <= n ; i++)

87         {

88             ans = min(ans,solve(p[i]));

89         }

90         printf("%.4f\n",ans);

91     }

92     return 0;

93 }

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