简单多边形与圆交面积模板
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简单多边形与圆交面积模板
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1 /*
2 求多邊形與圓相交的面積
3 */
4
5 #include <iostream>
6 #include <cstdio>
7 #include <cmath>
8 #define max(x,y) ((x)>(y)?(x):(y))
9 #define min(x,y) ((x)<(y)?(x):(y))
10 #define PI acos(-1.0)
11 #define EPS 1e-10
12 using namespace std;
13
14 inline int DB(double x) {
15 if(x<-EPS) return -1;
16 if(x>EPS) return 1;
17 return 0;
18 }
19
20
21 struct point {
22 double x,y;
23
24 point() {}
25 point(double _x,double _y):x(_x),y(_y) {}
26
27 point operator-(point a) {
28 return point(x-a.x,y-a.y);
29 }
30 point operator+(point a) {
31 return point(x+a.x,y+a.y);
32 }
33
34 double operator*(point a) {
35 return x*a.y-y*a.x;
36 }
37
38 point oppose() {
39 return point(-x,-y);
40 }
41
42 double length() {
43 return sqrt(x*x+y*y);
44 }
45
46 point adjust(double L) {
47 L/=length();
48 return point(x*L,y*L);
49 }
50
51 point vertical() {
52 return point(-y,x);
53 }
54
55 double operator^(point a) {
56 return x*a.x+y*a.y;
57 }
58 };
59
60
61 struct segment {
62 point a,b;
63
64 segment() {}
65 segment(point _a,point _b):a(_a),b(_b) {}
66
67 point intersect(segment s) {
68 double s1=(s.a-a)*(s.b-a);
69 double s2=(s.b-b)*(s.a-b);
70 double t=s1+s2;
71 s1/=t;
72 s2/=t;
73 return point(a.x*s2+b.x*s1,a.y*s2+b.y*s1);
74 }
75 point vertical(point p) {
76 point t=(b-a).vertical();
77 return intersect(segment(p,p+t));
78 }
79 int isonsegment(point p) {
80 return DB(min(a.x,b.x)-p.x)<=0&&
81 DB(max(a.x,b.x)-p.x)>=0&&
82 DB(min(a.y,b.y)-p.y)<=0&&
83 DB(max(a.y,b.y)-p.y)>=0;
84 }
85 };
86
87 struct circle {
88 point p;
89 double R;
90 };
91
92 circle C;
93 point p[105];
94 int n;
95
96
97 double cross_area(point a,point b,circle C) {
98 point p=C.p;
99 double R=C.R;
100 int sgn=DB((b-p)*(a-p));
101 double La=(a-p).length(),Lb=(b-p).length();
102 int ra=DB(La-R),rb=DB(Lb-R);
103 double ang=acos(((b-p)^(a-p))/(La*Lb));
104 segment t(a,b);
105 point s;
106 point h,u,temp;
107 double ans,L,d,ang1;
108
109 if(!DB(La)||!DB(Lb)||!sgn) ans=0;
110 else if(ra<=0&&rb<=0) ans=fabs((b-p)*(a-p))/2;
111 else if(ra>=0&&rb>=0) {
112 h=t.vertical(p);
113 L=(h-p).length();
114 if(!t.isonsegment(h)||DB(L-R)>=0) ans=R*R*(ang/2);
115 else {
116 ans=R*R*(ang/2);
117 ang1=acos(L/R);
118 ans-=R*R*ang1;
119 ans+=R*sin(ang1)*L;
120 }
121 } else {
122 h=t.vertical(p);
123 L=(h-p).length();
124 s=b-a;
125 d=sqrt(R*R-L*L);
126 s=s.adjust(d);
127 if(t.isonsegment(h+s)) u=h+s;
128 else u=h+s.oppose();
129 if(ra==1) temp=a,a=b,b=temp;
130 ans=fabs((a-p)*(u-p))/2;
131 ang1=acos(((u-p)^(b-p))/((u-p).length()*(b-p).length()));
132 ans+=R*R*(ang1/2);
133 }
134 return ans*sgn;
135 }
136
137 double cal_cross(circle C,point p[],int n) {
138 double ans=0;
139 int i;
140 p[n]=p[0];
141 for(i=0; i<n; i++) ans+=cross_area(p[i],p[i+1],C);
142 return fabs(ans);
143 }
144
145 double x,y,v,det,t,g,R;
146
147 int input() {
148 scanf("%lf%lf%lf%lf%lf%lf%lf",&x,&y,&v,&det,&t,&g,&R);
149 return x||y||v||det||t||g||R;
150 }
151
152
153 int main() {
154 //freopen("test.in","r",stdin);
155 while(input()) {
156 scanf("%d",&n);
157 int i;
158 for(i=0; i<n; i++) scanf("%lf%lf",&p[i].x,&p[i].y);
159 det=det/180*PI;
160 C.R=R;
161 C.p.x=x+v*cos(det)*t;
162 C.p.y=y+v*sin(det)*t-0.5*g*t*t;
163 double area=cal_cross(C,p,n);
164 printf("%.2lf\n",area);
165 }
166 return 0;
167 } View Code
轉自：http://blog.csdn.net/acm_baihuzi/article/details/48473725
多邊形面積求法：https://www.cnblogs.com/xiexinxinlove/p/3708147.html
轉載于:https://www.cnblogs.com/zmin/p/8329118.html
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