Poj--3831（两圆相交，二分）

最新推荐文章于 2021-03-18 22:30:38 发布

转载最新推荐文章于 2021-03-18 22:30:38 发布 · 105 阅读

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原文链接：http://www.cnblogs.com/naturepengchen/articles/4139026.html

本文介绍了解决广州热身赛C题的方法，通过枚举大圆圆心并使用二分查找确定大圆半径的方式，确保覆盖所有小圆。代码中详细展示了如何计算两圆相交面积，并验证大圆是否能完全覆盖每个小圆。

2014-12-03 01:32:33

思路：广州热身赛C题的原题，枚举大圆圆心，二分大圆半径。

 1 /*************************************************************************
 2     > File Name: p3831.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com
 5     > Created Time: Wed 03 Dec 2014 12:50:01 AM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <vector>
13 #include <map>
14 #include <set>
15 #include <stack>
16 #include <queue>
17 #include <iostream>
18 #include <algorithm>
19 using namespace std;
20 #define lp (p << 1)
21 #define rp (p << 1|1)
22 #define getmid(l,r) (l + (r - l) / 2)
23 #define MP(a,b) make_pair(a,b)
24 typedef long long ll;
25 const int INF = 1 << 30;
26 const double PI = acos(-1.0);
27 const int RA = 1e9;
28 const double eps = 1e-8;
29 
30 int T;
31 int N;
32 double X[30],Y[30],R[30];
33 
34 double Cal(double ax,double ay,double ar,double bx,double by,double br){
35     double d = sqrt((ax - bx) * (ax - bx) + (ay - by) * (ay - by));
36     if(d >= ar + br)
37         return 0.0;
38     double r = min(ar,br);
39     if(d <= fabs(ar - br))
40         return PI * r * r;
41     double a1 = acos((ar * ar + d * d - br * br) / (2.0 * ar * d));
42     double a2 = acos((br * br + d * d - ar * ar) / (2.0 * br * d));
43     double ans = -d * ar * sin(a1);
44     ans += a1 * ar * ar + a2 * br * br;
45     return ans;
46 }
47 
48 bool Check(int c,double r){
49     for(int i = 1; i <= N; ++i){
50         double s = PI * R[i] * R[i];
51         double its = 2.0 * Cal(X[c],Y[c],r,X[i],Y[i],R[i]);
52         if(its < s)
53             return false;
54     }
55     return true;
56 }
57 
58 int main(){
59     double a,b,r;
60     scanf("%d",&T);
61     while(T--){
62         scanf("%d",&N);
63         for(int i = 1; i <= N; ++i){
64             scanf("%lf%lf%lf",&X[i],&Y[i],&R[i]);
65         }
66         double ans = INF;
67         for(int i = 1; i <= N; ++i){
68             double l = 0,r = RA;
69             while(fabs(r - l) >= eps){
70                 double mid = getmid(l,r);
71                 if(Check(i,mid))
72                     r = mid;
73                 else
74                     l = mid;
75             }
76             ans = min(ans,l);
77         }
78         printf("%.4f\n",ans);
79     }
80     return 0;
81 }

转载于:https://www.cnblogs.com/naturepengchen/articles/4139026.html