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hdu 3264 Open-air shopping malls(求圆相交的面积,二分)

时间:2015-11-06 16:07:32      阅读:205      评论:0      收藏:0      [点我收藏+]

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Open-air shopping malls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2256    Accepted Submission(s): 837


Problem Description
The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping.

Unfortunately, the climate has changed little by little and now rainy days seriously affected the operation of open-air shopping malls—it’s obvious that nobody will have a good mood when shopping in the rain. In order to change this situation, the manager of these open-air shopping malls would like to build a giant umbrella to solve this problem. 

These shopping malls can be considered as different circles. It is guaranteed that these circles will not intersect with each other and no circles will be contained in another one. The giant umbrella is also a circle. Due to some technical reasons, the center of the umbrella must coincide with the center of a shopping mall. Furthermore, a fine survey shows that for any mall, covering half of its area is enough for people to seek shelter from the rain, so the task is to decide the minimum radius of the giant umbrella so that for every shopping mall, the umbrella can cover at least half area of the mall.
 

 

Input
The input consists of multiple test cases. 
The first line of the input contains one integer T (1<=T<=10), which is the number of test cases.
For each test case, there is one integer N (1<=N<=20) in the first line, representing the number of shopping malls.
The following N lines each contain three integers X,Y,R, representing that the mall has a shape of a circle with radius R and its center is positioned at (X,Y). X and Y are in the range of [-10000,10000] and R is a positive integer less than 2000.
 

 

Output
For each test case, output one line contains a real number rounded to 4 decimal places, representing the minimum radius of the giant umbrella that meets the demands.
 

 

Sample Input
1 2 0 0 1 2 0 1
 

 

Sample Output
2.0822
 

 

Source
 

 

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lcy   |   We have carefully selected several similar problems for you:  3268 3265 3269 3262 3263 
 
 
主要是求圆相交的面积。。
画图易得。
然后二分半径就好。
技术分享
  1 /*************************************************************************
  2     > File Name: code/hdu/3264.cpp
  3     > Author: 111qqz
  4     > Email: rkz2013@126.com 
  5     > Created Time: 2015年11月06日 星期五 14时57分47秒
  6  ************************************************************************/
  7 
  8 #include<iostream>
  9 #include<iomanip>
 10 #include<cstdio>
 11 #include<algorithm>
 12 #include<cmath>
 13 #include<cstring>
 14 #include<string>
 15 #include<map>
 16 #include<set>
 17 #include<queue>
 18 #include<vector>
 19 #include<stack>
 20 #include<cctype>
 21 #define fst first              
 22 #define sec second      
 23 #define lson l,m,rt<<1
 24 #define rson m+1,r,rt<<1|1
 25 #define ms(a,x) memset(a,x,sizeof(a))
 26 using namespace std;
 27 const double eps = 1E-8;
 28 const int dx4[4]={1,0,0,-1};
 29 const int dy4[4]={0,-1,1,0};
 30 typedef long long LL;
 31 const int inf = 0x3f3f3f3f;
 32 const int N=25;
 33 const double pi =acos(-1.0);
 34 int n;
 35 int dblcmp(double d)
 36 {
 37     return d<-eps?-1:d>eps;
 38 }
 39 struct point
 40 {
 41     double x,y;
 42     point(){}
 43     point(double _x,double _y):
 44     x(_x),y(-y){};
 45     void input()
 46     {
 47     scanf("%lf%lf",&x,&y);
 48     }
 49     point sub(point p)
 50     {
 51     return point(x-p.x,y-p.y);
 52     }
 53     double dot(point p)
 54     {
 55     return x*p.x+y*p.y;
 56     }
 57     double det(point p)
 58     {
 59     return x*p.y-y*p.x;
 60     }
 61     double distance(point p)
 62     {
 63     return hypot(x-p.x,y-p.y);
 64     }
 65 };
 66 
 67 struct circle
 68 {
 69     point p;
 70     double r;
 71     circle(){}
 72     circle(point _p,double _r):
 73     p(_p),r(_r){};
 74     void input()
 75     {
 76     p.input();
 77     scanf("%lf",&r);
 78     }
 79     double area()
 80     {
 81     return pi*r*r;
 82     }
 83 
 84     int relationcircle(circle v)
 85     {
 86     double d=p.distance(v.p);
 87     if (dblcmp(d-r-v.r)>0) return 5;
 88     if (dblcmp(d-r-v.r)==0) return 4;
 89     double l = fabs(r-v.r);
 90     if (dblcmp(d-r-v.r)<0&&dblcmp(d-l)>0) return 3;
 91     if (dblcmp(d-l)==0) return 2;
 92     if (dblcmp(d-l)<0) return 1;
 93     }
 94 
 95     double areacircle(circle v)
 96     {
 97     int rel = relationcircle(v);
 98     if (rel>=4) return 0.0;
 99     if (rel<=2) return min(area(),v.area());
100     double d=p.distance(v.p);
101     double hf=(r+v.r+d)/2.0;
102     double ss=2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
103     double a1=acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
104     a1 = a1 *r*r;
105     double a2=acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
106     a2 = a2*v.r*v.r;
107     return a1+a2-ss;
108     }
109 }cir[N];
110 
111 bool judge(double rr)
112 {
113     for ( int i = 1 ,j; i <= n ; i++)
114     {
115     circle tmp = cir[i];
116     tmp.r = rr;
117     for (  j = 1; j <= n ; j++)
118     {
119         double crossarea = tmp.areacircle(cir[j]);
120        // cout<<"cross_area:"<<crossarea<<endl;
121         if (crossarea<cir[j].area()/2.0)
122         break;
123     }
124     if (j==n+1) return true;
125     
126     }
127     return false;
128 
129 }
130 int main()
131 {
132   #ifndef  ONLINE_JUDGE 
133    freopen("in.txt","r",stdin);
134   #endif
135    int T;
136    scanf("%d",&T);
137    while ( T-- )
138     {
139     scanf("%d",&n);
140     for ( int i = 1  ; i <=  n ; i++) cir[i].input();
141     double l=0,r=1E5;
142     while (dblcmp(r-l)>0)
143     {
144        // cout<<"l:"<<l<<" r:"<<r<<endl;
145         double mid = (l+r)/2.0;
146         if (judge(mid))
147         r = mid;
148         else l=mid;
149 
150     }
151     printf("%.4f\n",r);
152 
153     }
154   
155    
156  #ifndef ONLINE_JUDGE  
157   #endif
158   fclose(stdin);
159     return 0;
160 }
View Code

 

hdu 3264 Open-air shopping malls(求圆相交的面积,二分)

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原文地址:http://www.cnblogs.com/111qqz/p/4942802.html

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