HDU 3264/POJ 3831 Open-air shopping malls（计算几何+二分）（2009 Asia Ningbo Regional）

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.

题目大意：给n个互相相离的圆，要求以其中一个圆的圆点为中心，画一个大圆，这个大圆要覆盖这n个圆至少一半的面积，求这个大圆的最小半径。

思路：枚举圆心，二分半径判断是否符合条件即可。

代码（0MS）：

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <cmath>
 using namespace std;

 const int MAXN = ;
 const double PI =  * acos(0.0);
 const double EPF = 1e-;

 double x[MAXN], y[MAXN], r[MAXN];
 int n, T;

 double dist(double x1, double y1, double x2, double y2) {
     int xx = x1 - x2, yy = y1 - y2;
     return sqrt(xx * xx + yy * yy);
 }

 double fusiform(double a, double c, double b) {
     double angle =  * acos((a * a + b * b - c * c) / ( * a * b));
     double s1 = a * a * PI * (angle / ( * PI));
     double s2 = a * a * sin(angle) / ;
     return s1 - s2;
 }

 bool common(double x1, double y1, double r1, double x2, double y2, double r2) {
     double d = dist(x1, y1, x2, y2);
     if(d >= r1 + r2) return false;
     if(d <= fabs(r1 - r2)) {
         if(r1 > r2) return true;
         else return (r1 * r1 *  >= r2 * r2);
     }
     double value = fusiform(r1, r2, d) + fusiform(r2, r1, d);
     return value *  >= r2 * r2 * PI;
 }

 bool check(double ox, double oy, double rr) {
     for(int i = ; i <= n; ++i)
         if(!common(ox, oy, rr, x[i], y[i], r[i])) return false;
     return true;
 }

 double calc(double ox, double oy) {
     double l = , r = ;
     while(r - l > 1e-) {
         double mid = (l + r) / ;
         if(check(ox, oy, mid)) r = mid;
         else l = mid;
     }
     return l;
 }

 void solve() {
     double ans = 1e100, value;
     for(int i = ; i <= n; ++i) {
         value = calc(x[i], y[i]);
         ans = min(ans, value);
     }
     printf("%.4f\n", ans);
 }

 int main() {
     //cout<<PI<<endl;
     scanf("%d", &T);
     while(T--) {
         scanf("%d", &n);
         for(int i = ; i <= n; ++i) scanf("%lf%lf%lf", &x[i], &y[i], &r[i]);
         solve();
     }
 }