Open-air shopping malls
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2344 Accepted Submission(s): 866
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.
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.
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
题意：平面上有一些圆,现在找一个圆的圆心为圆心做一个大圆出来，然后这个大圆至少要包含每个圆的1/2的面积，求这个大圆的最小半径。
题解：枚举每个点的圆心，二分求解。。WA了好久，竟然是我的圆公共面积的模板有问题，，幸好查出来了，，不然以后做模板的话有问题都不知道错哪里。。
内含的double 习惯性的打成了int
#include<stdio.h> #include<iostream> #include<string.h> #include <stdlib.h> #include<math.h> #include<algorithm> using namespace std; const int N = 25; const double pi = atan(1.0)*4; const double eps = 1e-8; struct Circle{ double x,y,r; }c[N]; typedef Circle Point; double dis(Point a, Point b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); } ///两圆相交面积模板 double Two_Circle_Area(Circle a,Circle b){ double d = dis(a,b); if(a.r+b.r<d){ ///相离 return 0; } if(fabs(a.r-b.r)>=d){ ///内含 double r = min(a.r,b.r); return pi*r*r; } double angleA = acos((a.r*a.r+d*d-b.r*b.r)/2/a.r/d); double angleB = acos((b.r*b.r+d*d-a.r*a.r)/2/b.r/d); double area1 = a.r*a.r*angleA; double area2 = b.r*b.r*angleB; return area1+area2-a.r*d*sin(angleA); } int n; double binary(double l,double r,Circle circle){ double mid; while((r-l)>=eps){ mid = (l+r)/2; circle.r = mid; bool flag = false; for(int i=0;i<n;i++){ double area = pi*c[i].r*c[i].r; if(area/2>Two_Circle_Area(circle,c[i])){ flag = true; break; } } if(flag) l =mid; else r = mid; } return mid; } int main(){ int tcase; scanf("%d",&tcase); while(tcase--){ scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%lf%lf%lf",&c[i].x,&c[i].y,&c[i].r); } Circle circle; double mi = 99999999999; for(int i=0;i<n;i++){ circle.x = c[i].x; circle.y = c[i].y; double l=0,r = 20000; mi = min(mi,binary(l,r,circle)); } printf("%.4lf ",mi); } return 0; }