HDU 5120 Intersection （求圆环相交面积）

Posted 2020-09-30 一个人的时光

题意：给定圆环的内径r和外径R，以及2个相同圆环的圆心，求两个圆环的相交面积。

思路：

S = A大B大 - A大B小 - A小B大 + A小B小。（A表示A环，大表示大圆，B同）

 1 #include <iostream>
 2 #include <queue>
 3 #include <stack>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <map>
 7 #include <set>
 8 #include <bitset>
 9 #include <algorithm>
10 #include <cmath>
11 #include <cstring>
12 #include <cstdlib>
13 #include <string>
14 #include <sstream>
15 #include <time.h>
16 #define x first
17 #define y second
18 #define pb push_back
19 #define mp make_pair
20 #define lson l,m,rt*2
21 #define rson m+1,r,rt*2+1
22 #define mt(A,B) memset(A,B,sizeof(A))
23 using namespace std;
24 typedef long long LL;
25 typedef unsigned long long ull;
26 const double PI = acos(-1);
27 const int N=1e6+10;
28 const LL mod=1e9+7;
29 const int inf = 0x3f3f3f3f;
30 const LL INF=0x3f3f3f3f3f3f3f3fLL;
31 const double eps=1e-8;
32 struct point
33 {
34     double x,y;
35     point(){}
36     point(double _x,double _y)
37     {
38         x=_x;
39         y=_y;
40     }
41 };
42 double dis(point a,point b)
43 {
44     return sqrt((a.x-b.x)*(a.x-b.x)+(b.y-a.y)*(b.y-a.y));
45 }
46 double cal(point c1,double r1,point c2,double r2)
47 {
48     double d=dis(c1,c2);
49     if(r1+r2<d+eps)return 0;
50     if(d<fabs(r1-r2)+eps)
51     {
52         double r=min(r1,r2);
53         return PI*r*r;
54     }
55     double x=(d*d+r1*r1-r2*r2)/(2*d);
56     double t1=acos(x/r1);
57     double t2=acos((d-x)/r2);
58     return r1*r1*t1+r2*r2*t2-d*r1*sin(t1);
59 }
60 int main()
61 {
62 #ifdef Local
63     freopen("data.h","r",stdin);
64 #endif
65     //ios::sync_with_stdio(false);
66     //cin.tie(0);
67     int cas=1,T,n,m;
68     scanf("%d",&T);
69     while(T--)
70     {
71         double r,R,ans;
72         struct point p,q;
73         cin>>r>>R;
74         cin>>p.x>>p.y;
75         cin>>q.x>>q.y;
76         if(dis(p,q)>=2*R)ans=0;
77         else
78         {
79             if(dis(p,q)>=2*r)
80             {
81                 ans=cal(p,R,q,R)-cal(p,r,q,R)-cal(q,r,p,R);
82             }
83             else
84             {
85                 ans=cal(p,R,q,R)-cal(p,r,q,R)-cal(q,r,p,R)+cal(q,r,p,r);
86             }
87         }
88         printf("Case #%d: %lf\\n",cas++,ans);
89     }
90 #ifdef Local
91     cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
92 #endif
93    return 0;
94 }

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