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  • 求两个圆的重合面积+二分 hdu3264

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <cstring>
      4 #include <algorithm>
      5 #include <cmath>
      6 
      7 using namespace std;
      8 
      9 struct circle
     10 {
     11     double x,y,r;
     12 }c[25];
     13 
     14 double exp1 = 10e-9;
     15 double reans[25];
     16 
     17 double iabs(double a)
     18 {
     19     if(a<0)
     20         return -a;
     21     else
     22         return a;
     23 }
     24 
     25 double mianji(circle c1)
     26 {
     27     return c1.r*c1.r*acos(-1);
     28 }
     29 
     30 double jieguo(circle c1,circle c2,double r)
     31 {
     32     double b = sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
     33     double c = c2.r;
     34     double a = r;
     35     if(a+c<=b)
     36         return 0;
     37     if(b<=iabs(a-c))
     38     {
     39         return min(a,c)*min(a,c)*acos(-1);
     40     }
     41     double cos1 = (b*b+c*c-a*a)/2/c/b;
     42     double s1 = c*c*acos(cos1);
     43     double cos2 = (a*a+b*b-c*c)/2/a/b;
     44     double s2 = a*a*acos(cos2);
     45     double s3 = a*b*sin(acos(cos2));
     46     double s4=s1+s2-s3;
     47     return s4;
     48 }
     49 
     50 int main()
     51 {
     52     int T;
     53     scanf("%d",&T);
     54     while(T--)
     55     {
     56         int n;
     57         scanf("%d",&n);
     58         for(int i=0;i<n;i++)
     59         {
     60             scanf("%lf%lf%lf",&c[i].x,&c[i].y,&c[i].r);
     61         }
     62         memset(reans,0,sizeof(reans));
     63         for(int i=0;i<n;i++)
     64         {
     65             for(int t=0;t<n;t++)
     66             {
     67                 double s=c[t].r*c[t].r*acos(-1)/2;
     68                 double mar=sqrt((c[i].x-c[t].x)*(c[i].x-c[t].x)+(c[i].y-c[t].y)*(c[i].y-c[t].y))+c[t].r;
     69                 double mir=sqrt((c[i].x-c[t].x)*(c[i].x-c[t].x)+(c[i].y-c[t].y)*(c[i].y-c[t].y));
     70                 double ans=0;
     71                 double r=mir;
     72                 while(iabs(ans-s)>exp1)
     73                 {
     74                     if(ans<s)
     75                     {
     76                         mir=r;
     77                         r=(mir+mar)/2;
     78                         ans=jieguo(c[i],c[t],r);
     79                     }
     80                     else
     81                     {
     82                         mar=r;
     83                         r=(mir+mar)/2;
     84                         ans=jieguo(c[i],c[t],r);
     85                     }
     86                 }
     87                 if(reans[i]<r)
     88                     reans[i]=r;
     89             }
     90         }
     91         double ss=reans[0];
     92         for(int i=1;i<n;i++)
     93         {
     94             if(reans[i]<ss)
     95                 ss=reans[i];
     96         }
     97         printf("%.4lf
    ",ss);
     98     }
     99     return 0;
    100 }
    View Code
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  • 原文地址:https://www.cnblogs.com/wsruning/p/4853108.html
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