题意：有一块蛋糕，上面有一颗cherry。用刀子切n次，求切完之后有cherry的那部分的面积
My solution：
先做一个大矩形，使cake内切于这个大矩形。如图：
然后不断切这个大矩形，每次切割的时候保留与cherry同侧的那部分。最后剩下的就是一个多边形。求该多边形与圆的面积交即可。
在切割的时候如何保证留下来的是与cherry同侧的部分呢？很简单
方法不难，但是一直WA= =。遇到了个奇怪的问题：
对于这组数据：
3
5 2
-5 0 5 3
-5 0 5 -3
0 0
5 2
-5 0 5 3
-5 0 5 -3
0 4.9
5 2
-5 0 5 3
-5 0 5 -3
0 -4.9
画出来图是这样的：
标程输出结果：
My solution：
可是标程明显不对啊尼玛！加起来都超过1了是什么鬼！
思考ing.........
附WA code：
1 #include<vector>
2 #include<list>
3 #include<map>
4 #include<set>
5 #include<deque>
6 #include<queue>
7 #include<stack>
8 #include<bitset>
9 #include<algorithm>
10 #include<functional>
11 #include<numeric>
12 #include<utility>
13 #include<iostream>
14 #include<sstream>
15 #include<iomanip>
16 #include<cstdio>
17 #include<cmath>
18 #include<cstdlib>
19 #include<cctype>
20 #include<string>
21 #include<cstring>
22 #include<cstdio>
23 #include<cmath>
24 #include<cstdlib>
25 #include<ctime>
26 #include<climits>
27 #include<complex>
28 #define mp make_pair
29 #define pb push_back
30 using namespace std;
31 const double eps=1e-6;
32 const double pi=acos(-1.0);
33 const double inf=1e20;
34 const int maxp=11000;
35
36 int sgn(double x)
37 {
38 if (fabs(x)<eps) return 0;
39 if (x<0) return -1;
40 else return 1;
41 }
42
43 int dblcmp(double d)
44 {
45 if (fabs(d)<eps)return 0;
46 return d>eps?1:-1;
47 }
48
49 inline double sqr(double x){return x*x;}
50
51 struct point
52 {
53 double x,y;
54 point(){}
55 point(double _x,double _y):
56 x(_x),y(_y){};
57 void input()
58 {
59 scanf("%lf%lf",&x,&y);
60 }
61 void output()
62 {
63 printf("%.2f %.2f\n",x,y);
64 }
65 bool operator==(point a)const
66 {
67 return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
68 }
69 bool operator<(point a)const
70 {
71 return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
72 }
73
74 point operator +(const point &b)const
75 {
76 return point(x+b.x,y+b.y);
77 }
78 point operator -(const point &b)const
79 {
80 return point(x-b.x,y-b.y);
81 }
82 point operator *(const double &k)const
83 {
84 return point(x*k,y*k);
85 }
86 point operator /(const double &k)const
87 {
88 return point(x/k,y/k);
89 }
90 double operator *(const point &b)const
91 {
92 return x*b.x+y*b.y;
93 }
94 double operator ^(const point &b)const
95 {
96 return x*b.y-y*b.x;
97 }
98
99 double len()
100 {
101 return hypot(x,y);
102 }
103 double len2()
104 {
105 return x*x+y*y;
106 }
107 double distance(point p)
108 {
109 return hypot(x-p.x,y-p.y);
110 }
111 point add(point p)
112 {
113 return point(x+p.x,y+p.y);
114 }
115 point sub(point p)
116 {
117 return point(x-p.x,y-p.y);
118 }
119 point mul(double b)
120 {
121 return point(x*b,y*b);
122 }
123 point div(double b)
124 {
125 return point(x/b,y/b);
126 }
127 double dot(point p)
128 {
129 return x*p.x+y*p.y;
130 }
131 double det(point p)
132 {
133 return x*p.y-y*p.x;
134 }
135 double rad(point a,point b)
136 {
137 point p=*this;
138 return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
139 }
140 point trunc(double r)
141 {
142 double l=len();
143 if (!dblcmp(l))return *this;
144 r/=l;
145 return point(x*r,y*r);
146 }
147 point rotleft()
148 {
149 return point(-y,x);
150 }
151 point rotright()
152 {
153 return point(y,-x);
154 }
155 point rotate(point p,double angle)//绕点p逆时针旋转angle角度
156 {
157 point v=this->sub(p);
158 double c=cos(angle),s=sin(angle);
159 return point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
160 }
161 };
162
163 struct line
164 {
165 point a,b;
166 line(){}
167 line(point _a,point _b)
168 {
169 a=_a;
170 b=_b;
171 }
172 bool operator==(line v)
173 {
174 return (a==v.a)&&(b==v.b);
175 }
176 //倾斜角angle
177 line(point p,double angle)
178 {
179 a=p;
180 if (dblcmp(angle-pi/2)==0)
181 {
182 b=a.add(point(0,1));
183 }
184 else
185 {
186 b=a.add(point(1,tan(angle)));
187 }
188 }
189 //ax+by+c=0
190 line(double _a,double _b,double _c)
191 {
192 if (dblcmp(_a)==0)
193 {
194 a=point(0,-_c/_b);
195 b=point(1,-_c/_b);
196 }
197 else if (dblcmp(_b)==0)
198 {
199 a=point(-_c/_a,0);
200 b=point(-_c/_a,1);
201 }
202 else
203 {
204 a=point(0,-_c/_b);
205 b=point(1,(-_c-_a)/_b);
206 }
207 }
208 void input()
209 {
210 a.input();
211 b.input();
212 }
213 void adjust()
214 {
215 if (b<a)swap(a,b);
216 }
217 double length()
218 {
219 return a.distance(b);
220 }
221 double angle()//直线倾斜角 0<=angle<180
222 {
223 double k=atan2(b.y-a.y,b.x-a.x);
224 if (dblcmp(k)<0)k+=pi;
225 if (dblcmp(k-pi)==0)k-=pi;
226 return k;
227 }
228 //点和线段关系
229 //1 在逆时针
230 //2 在顺时针
231 //3 平行
232 int relation(point p)
233 {
234 int c=dblcmp(p.sub(a).det(b.sub(a)));
235 if (c<0)return 1;
236 if (c>0)return 2;
237 return 3;
238 }
239 bool pointonseg(point p)
240 {
241 return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
242 }
243 bool parallel(line v)
244 {
245 return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
246 }
247 //2 规范相交
248 //1 非规范相交
249 //0 不相交
250 int segcrossseg(line v)
251 {
252 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
253 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
254 int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
255 int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
256 if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
257 return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
258 d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
259 d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
260 d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
261 }
262 int linecrossseg(line v)//*this seg v line
263 {
264 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
265 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
266 if ((d1^d2)==-2)return 2;
267 return (d1==0||d2==0);
268 }
269 //0 平行
270 //1 重合
271 //2 相交
272 int linecrossline(line v)
273 {
274 if ((*this).parallel(v))
275 {
276 return v.relation(a)==3;
277 }
278 return 2;
279 }
280 point crosspoint(line v)
281 {
282 double a1=v.b.sub(v.a).det(a.sub(v.a));
283 double a2=v.b.sub(v.a).det(b.sub(v.a));
284 return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
285 }
286 double dispointtoline(point p)
287 {
288 return fabs(p.sub(a).det(b.sub(a)))/length();
289 }
290 double dispointtoseg(point p)
291 {
292 if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
293 {
294 return min(p.distance(a),p.distance(b));
295 }
296 return dispointtoline(p);
297 }
298 point lineprog(point p)
299 {
300 return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a))/b.sub(a).len2()));
301 }
302 point symmetrypoint(point p)
303 {
304 point q=lineprog(p);
305 return point(2*q.x-p.x,2*q.y-p.y);
306 }
307 };
308
309 struct Vector:public point
310 {
311 Vector(){}
312 Vector(double a,double b)
313 {
314 x=a; y=b;
315 }
316 Vector(point _a,point _b) //a->b
317 {
318 double dx=_b.x-_a.x;
319 double dy=_b.y-_a.y;
320 x=dx; y=dy;
321 }
322 Vector(line v)
323 {
324 double dx=v.b.x-v.a.x;
325 double dy=v.b.y-v.a.y;
326 x=dx; y=dy;
327 }
328 double length()
329 {
330 return (sqrt(x*x+y*y));
331 }
332 Vector Normal()
333 {
334 double L=sqrt(x*x+y*y);
335 Vector Vans=Vector(-y/L,x/L);
336 return Vans;
337 }
338 };
339
340 struct circle
341 {
342 point p;
343 double r;
344 circle(){}
345 circle(point _p,double _r):
346 p(_p),r(_r){};
347 circle(double x,double y,double _r):
348 p(point(x,y)),r(_r){};
349 circle(point a,point b,point c)//三角形的外接圆
350 {
351 p=line(a.add(b).div(2),a.add(b).div(2).add(b.sub(a).rotleft())).crosspoint(line(c.add(b).div(2),c.add(b).div(2).add(b.sub(c).rotleft())));
352 r=p.distance(a);
353 }
354 circle(point a,point b,point c,bool t)//三角形的内切圆
355 {
356 line u,v;
357 double m=atan2(b.y-a.y,b.x-a.x),n=atan2(c.y-a.y,c.x-a.x);
358 u.a=a;
359 u.b=u.a.add(point(cos((n+m)/2),sin((n+m)/2)));
360 v.a=b;
361 m=atan2(a.y-b.y,a.x-b.x),n=atan2(c.y-b.y,c.x-b.x);
362 v.b=v.a.add(point(cos((n+m)/2),sin((n+m)/2)));
363 p=u.crosspoint(v);
364 r=line(a,b).dispointtoseg(p);
365 }
366 void input()
367 {
368 p.input();
369 scanf("%lf",&r);
370 }
371 void output()
372 {
373 printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);
374 }
375 bool operator==(circle v)
376 {
377 return ((p==v.p)&&dblcmp(r-v.r)==0);
378 }
379 bool operator<(circle v)const
380 {
381 return ((p<v.p)||(p==v.p)&&dblcmp(r-v.r)<0);
382 }
383 double area()
384 {
385 return pi*sqr(r);
386 }
387 double circumference()
388 {
389 return 2*pi*r;
390 }
391 //0 圆外
392 //1 圆上
393 //2 圆内
394 int relation(point b)
395 {
396 double dst=b.distance(p);
397 if (dblcmp(dst-r)<0)return 2;
398 if (dblcmp(dst-r)==0)return 1;
399 return 0;
400 }
401 int relationseg(line v)
402 {
403 double dst=v.dispointtoseg(p);
404 if (dblcmp(dst-r)<0)return 2;
405 if (dblcmp(dst-r)==0)return 1;
406 return 0;
407 }
408 int relationline(line v)
409 {
410 double dst=v.dispointtoline(p);
411 if (dblcmp(dst-r)<0)return 2;
412 if (dblcmp(dst-r)==0)return 1;
413 return 0;
414 }
415 //过a b两点 半径r的两个圆
416 int getcircle(point a,point b,double r,circle&c1,circle&c2)
417 {
418 circle x(a,r),y(b,r);
419 int t=x.pointcrosscircle(y,c1.p,c2.p);
420 if (!t)return 0;
421 c1.r=c2.r=r;
422 return t;
423 }
424 //与直线u相切 过点q 半径r1的圆
425 int getcircle(line u,point q,double r1,circle &c1,circle &c2)
426 {
427 double dis=u.dispointtoline(q);
428 if (dblcmp(dis-r1*2)>0)return 0;
429 if (dblcmp(dis)==0)
430 {
431 c1.p=q.add(u.b.sub(u.a).rotleft().trunc(r1));
432 c2.p=q.add(u.b.sub(u.a).rotright().trunc(r1));
433 c1.r=c2.r=r1;
434 return 2;
435 }
436 line u1=line(u.a.add(u.b.sub(u.a).rotleft().trunc(r1)),u.b.add(u.b.sub(u.a).rotleft().trunc(r1)));
437 line u2=line(u.a.add(u.b.sub(u.a).rotright().trunc(r1)),u.b.add(u.b.sub(u.a).rotright().trunc(r1)));
438 circle cc=circle(q,r1);
439 point p1,p2;
440 if (!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);
441 c1=circle(p1,r1);
442 if (p1==p2)
443 {
444 c2=c1;return 1;
445 }
446 c2=circle(p2,r1);
447 return 2;
448 }
449 //同时与直线u,v相切 半径r1的圆
450 int getcircle(line u,line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4)
451 {
452 if (u.parallel(v))return 0;
453 line u1=line(u.a.add(u.b.sub(u.a).rotleft().trunc(r1)),u.b.add(u.b.sub(u.a).rotleft().trunc(r1)));
454 line u2=line(u.a.add(u.b.sub(u.a).rotright().trunc(r1)),u.b.add(u.b.sub(u.a).rotright().trunc(r1)));
455 line v1=line(v.a.add(v.b.sub(v.a).rotleft().trunc(r1)),v.b.add(v.b.sub(v.a).rotleft().trunc(r1)));
456 line v2=line(v.a.add(v.b.sub(v.a).rotright().trunc(r1)),v.b.add(v.b.sub(v.a).rotright().trunc(r1)));
457 c1.r=c2.r=c3.r=c4.r=r1;
458 c1.p=u1.crosspoint(v1);
459 c2.p=u1.crosspoint(v2);
460 c3.p=u2.crosspoint(v1);
461 c4.p=u2.crosspoint(v2);
462 return 4;
463 }
464 //同时与不相交圆cx,cy相切 半径为r1的圆
465 int getcircle(circle cx,circle cy,double r1,circle&c1,circle&c2)
466 {
467 circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);
468 int t=x.pointcrosscircle(y,c1.p,c2.p);
469 if (!t)return 0;
470 c1.r=c2.r=r1;
471 return t;
472 }
473 int pointcrossline(line v,point &p1,point &p2)//求与线段交要先判断relationseg
474 {
475 if (!(*this).relationline(v))return 0;
476 point a=v.lineprog(p);
477 double d=v.dispointtoline(p);
478 d=sqrt(r*r-d*d);
479 if (dblcmp(d)==0)
480 {
481 p1=a;
482 p2=a;
483 return 1;
484 }
485 p1=a.sub(v.b.sub(v.a).trunc(d));
486 p2=a.add(v.b.sub(v.a).trunc(d));
487 return 2;
488 }
489 //5 相离
490 //4 外切
491 //3 相交
492 //2 内切
493 //1 内含
494 int relationcircle(circle v)
495 {
496 double d=p.distance(v.p);
497 if (dblcmp(d-r-v.r)>0)return 5;
498 if (dblcmp(d-r-v.r)==0)return 4;
499 double l=fabs(r-v.r);
500 if (dblcmp(d-r-v.r)<0&&dblcmp(d-l)>0)return 3;
501 if (dblcmp(d-l)==0)return 2;
502 if (dblcmp(d-l)<0)return 1;
503 }
504 int pointcrosscircle(circle v,point &p1,point &p2)
505 {
506 int rel=relationcircle(v);
507 if (rel==1||rel==5)return 0;
508 double d=p.distance(v.p);
509 double l=(d+(sqr(r)-sqr(v.r))/d)/2;
510 double h=sqrt(sqr(r)-sqr(l));
511 p1=p.add(v.p.sub(p).trunc(l).add(v.p.sub(p).rotleft().trunc(h)));
512 p2=p.add(v.p.sub(p).trunc(l).add(v.p.sub(p).rotright().trunc(h)));
513 if (rel==2||rel==4)
514 {
515 return 1;
516 }
517 return 2;
518 }
519 //过一点做圆的切线 (先判断点和圆关系)
520 int tangentline(point q,line &u,line &v)
521 {
522 int x=relation(q);
523 if (x==2)return 0;
524 if (x==1)
525 {
526 u=line(q,q.add(q.sub(p).rotleft()));
527 v=u;
528 return 1;
529 }
530 double d=p.distance(q);
531 double l=sqr(r)/d;
532 double h=sqrt(sqr(r)-sqr(l));
533 u=line(q,p.add(q.sub(p).trunc(l).add(q.sub(p).rotleft().trunc(h))));
534 v=line(q,p.add(q.sub(p).trunc(l).add(q.sub(p).rotright().trunc(h))));
535 return 2;
536 }
537 double areacircle(circle v)
538 {
539 int rel=relationcircle(v);
540 if (rel>=4)return 0.0;
541 if (rel<=2)return min(area(),v.area());
542 double d=p.distance(v.p);
543 double hf=(r+v.r+d)/2.0;
544 double ss=2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));
545 double a1=acos((r*r+d*d-v.r*v.r)/(2.0*r*d));
546 a1=a1*r*r;
547 double a2=acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));
548 a2=a2*v.r*v.r;
549 return a1+a2-ss;
550 }
551 double areatriangle(point a,point b)
552 {
553 if (dblcmp(p.sub(a).det(p.sub(b))==0))return 0.0;
554 point q[5];
555 int len=0;
556 q[len++]=a;
557 line l(a,b);
558 point p1,p2;
559 if (pointcrossline(l,q[1],q[2])==2)
560 {
561 if (dblcmp(a.sub(q[1]).dot(b.sub(q[1])))<0)q[len++]=q[1];
562 if (dblcmp(a.sub(q[2]).dot(b.sub(q[2])))<0)q[len++]=q[2];
563 }
564 q[len++]=b;
565 if (len==4&&(dblcmp(q[0].sub(q[1]).dot(q[2].sub(q[1])))>0))swap(q[1],q[2]);
566 double res=0;
567 int i;
568 for (i=0;i<len-1;i++)
569 {
570 if (relation(q[i])==0||relation(q[i+1])==0)
571 {
572 double arg=p.rad(q[i],q[i+1]);
573 res+=r*r*arg/2.0;
574 }
575 else
576 {
577 res+=fabs(q[i].sub(p).det(q[i+1].sub(p))/2.0);
578 }
579 }
580 return res;
581 }
582 };
583
584 struct polygon
585 {
586 int n;
587 point p[maxp];
588 line l[maxp];
589 void input(int X)
590 {
591 n=X;
592 for (int i=0;i<n;i++)
593 {
594 p[i].input();
595 }
596 }
597 void add(point q)
598 {
599 p[n++]=q;
600 }
601 void getline()
602 {
603 for (int i=0;i<n;i++)
604 {
605 l[i]=line(p[i],p[(i+1)%n]);
606 }
607 }
608 struct cmp
609 {
610 point p;
611 cmp(const point &p0){p=p0;}
612 bool operator()(const point &aa,const point &bb)
613 {
614 point a=aa,b=bb;
615 int d=dblcmp(a.sub(p).det(b.sub(p)));
616 if (d==0)
617 {
618 return dblcmp(a.distance(p)-b.distance(p))<0;
619 }
620 return d>0;
621 }
622 };
623 void norm()
624 {
625 point mi=p[0];
626 for (int i=1;i<n;i++)mi=min(mi,p[i]);
627 sort(p,p+n,cmp(mi));
628 }
629 void getconvex(polygon &convex)
630 {
631 int i,j,k;
632 sort(p,p+n);
633 convex.n=n;
634 for (i=0;i<min(n,2);i++)
635 {
636 convex.p[i]=p[i];
637 }
638 if (n<=2)return;
639 int &top=convex.n;
640 top=1;
641 for (i=2;i<n;i++)
642 {
643 while (top&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
644 top--;
645 convex.p[++top]=p[i];
646 }
647 int temp=top;
648 convex.p[++top]=p[n-2];
649 for (i=n-3;i>=0;i--)
650 {
651 while (top!=temp&&convex.p[top].sub(p[i]).det(convex.p[top-1].sub(p[i]))<=0)
652 top--;
653 convex.p[++top]=p[i];
654 }
655 }
656
657 //ADD
658 //a new oonvex algorithm
659 /* void Graham(polygon &convex)
660 {
661 norm();
662 int &top=convex.n;
663 top=0;
664 if (n==1)
665 {
666 top=1;
667 convex.p[0]=p[0];
668 return;
669 }
670 if (n==2)
671 {
672 top=2;
673 convex.p[0]=p[0];
674 convex.p[1]=p[1];
675 if (convex.p[0]==convex.p[1]) top--;
676 return;
677 }
678 convex.p[0]=p[0];
679 convex.p[1]=p[1];
680 top=2;
681 for (int i=2;i<n;i++)
682 {
683 while (top>1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2]))<=0)
684 top--;
685 convex.p[top++]=p[i];
686 }
687 if (convex.n==2 && (convex.p[0]==convex.p[1])) convex.n--;
688 }
689 */
690 bool isconvex()
691 {
692 bool s[3];
693 memset(s,0,sizeof(s));
694 int i,j,k;
695 for (i=0;i<n;i++)
696 {
697 j=(i+1)%n;
698 k=(j+1)%n;
699 s[dblcmp(p[j].sub(p[i]).det(p[k].sub(p[i])))+1]=1;
700 if (s[0]&&s[2])return 0;
701 }
702 return 1;
703 }
704 //3 点上
705 //2 边上
706 //1 内部
707 //0 外部
708 int relationpoint(point q)
709 {
710 int i,j;
711 for (i=0;i<n;i++)
712 {
713 if (p[i]==q)return 3;
714 }
715 getline();
716 for (i=0;i<n;i++)
717 {
718 if (l[i].pointonseg(q))return 2;
719 }
720 int cnt=0;
721 for (i=0;i<n;i++)
722 {
723 j=(i+1)%n;
724 int k=dblcmp(q.sub(p[j]).det(p[i].sub(p[j])));
725 int u=dblcmp(p[i].y-q.y);
726 int v=dblcmp(p[j].y-q.y);
727 if (k>0&&u<0&&v>=0)cnt++;
728 if (k<0&&v<0&&u>=0)cnt--;
729 }
730 return cnt!=0;
731 }
732 //1 在多边形内长度为正
733 //2 相交或与边平行
734 //0 无任何交点
735 int relationline(line u)
736 {
737 int i,j,k=0;
738 getline();
739 for (i=0;i<n;i++)
740 {
741 if (l[i].segcrossseg(u)==2)return 1;
742 if (l[i].segcrossseg(u)==1)k=1;
743 }
744 if (!k)return 0;
745 vector<point>vp;
746 for (i=0;i<n;i++)
747 {
748 if (l[i].segcrossseg(u))
749 {
750 if (l[i].parallel(u))
751 {
752 vp.pb(u.a);
753 vp.pb(u.b);
754 vp.pb(l[i].a);
755 vp.pb(l[i].b);
756 continue;
757 }
758 vp.pb(l[i].crosspoint(u));
759 }
760 }
761 sort(vp.begin(),vp.end());
762 int sz=vp.size();
763 for (i=0;i<sz-1;i++)
764 {
765 point mid=vp[i].add(vp[i+1]).div(2);
766 if (relationpoint(mid)==1)return 1;
767 }
768 return 2;
769 }
770 //直线u切割凸多边形左侧
771 //注意直线方向
772 void convexcut(line u,polygon &po)
773 {
774 int i,j,k;
775 int &top=po.n;
776 top=0;
777 for (i=0;i<n;i++)
778 {
779 int d1=dblcmp(p[i].sub(u.a).det(u.b.sub(u.a)));
780 int d2=dblcmp(p[(i+1)%n].sub(u.a).det(u.b.sub(u.a)));
781 if (d1>=0)po.p[top++]=p[i];
782 if (d1*d2<0)po.p[top++]=u.crosspoint(line(p[i],p[(i+1)%n]));
783 }
784 }
785 double getcircumference()
786 {
787 double sum=0;
788 int i;
789 for (i=0;i<n;i++)
790 {
791 sum+=p[i].distance(p[(i+1)%n]);
792 }
793 return sum;
794 }
795 double getarea()
796 {
797 double sum=0;
798 int i;
799 for (i=0;i<n;i++)
800 {
801 sum+=p[i].det(p[(i+1)%n]);
802 }
803 return fabs(sum)/2;
804 }
805 bool getdir()//1代表逆时针 0代表顺时针
806 {
807 double sum=0;
808 int i;
809 for (i=0;i<n;i++)
810 {
811 sum+=p[i].det(p[(i+1)%n]);
812 }
813 if (dblcmp(sum)>0)return 1;
814 return 0;
815 }
816 point getbarycentre()
817 {
818 point ret(0,0);
819 double area=0;
820 int i;
821 for (i=1;i<n-1;i++)
822 {
823 double tmp=p[i].sub(p[0]).det(p[i+1].sub(p[0]));
824 if (dblcmp(tmp)==0)continue;
825 area+=tmp;
826 ret.x+=(p[0].x+p[i].x+p[i+1].x)/3*tmp;
827 ret.y+=(p[0].y+p[i].y+p[i+1].y)/3*tmp;
828 }
829 if (dblcmp(area))ret=ret.div(area);
830 return ret;
831 }
832 /* shen me gui !
833 double areaintersection(polygon po)
834 {
835 }
836 double areaunion(polygon po)
837 {
838 return getarea()+po.getarea()-areaintersection(po);
839 }
840 */
841 double areacircle(circle c)
842 {
843 int i,j,k,l,m;
844 double ans=0;
845 for (i=0;i<n;i++)
846 {
847 int j=(i+1)%n;
848 if (dblcmp(p[j].sub(c.p).det(p[i].sub(c.p)))>=0)
849 {
850 ans+=c.areatriangle(p[i],p[j]);
851 }
852 else
853 {
854 ans-=c.areatriangle(p[i],p[j]);
855 }
856 }
857 return fabs(ans);
858 }
859 //多边形和圆关系
860 //0 一部分在圆外
861 //1 与圆某条边相切
862 //2 完全在圆内
863 int relationcircle(circle c)
864 {
865 getline();
866 int i,x=2;
867 if (relationpoint(c.p)!=1)return 0;
868 for (i=0;i<n;i++)
869 {
870 if (c.relationseg(l[i])==2)return 0;
871 if (c.relationseg(l[i])==1)x=1;
872 }
873 return x;
874 }
875 void find(int st,point tri[],circle &c)
876 {
877 if (!st)
878 {
879 c=circle(point(0,0),-2);
880 }
881 if (st==1)
882 {
883 c=circle(tri[0],0);
884 }
885 if (st==2)
886 {
887 c=circle(tri[0].add(tri[1]).div(2),tri[0].distance(tri[1])/2.0);
888 }
889 if (st==3)
890 {
891 c=circle(tri[0],tri[1],tri[2]);
892 }
893 }
894 void solve(int cur,int st,point tri[],circle &c)
895 {
896 find(st,tri,c);
897 if (st==3)return;
898 int i;
899 for (i=0;i<cur;i++)
900 {
901 if (dblcmp(p[i].distance(c.p)-c.r)>0)
902 {
903 tri[st]=p[i];
904 solve(i,st+1,tri,c);
905 }
906 }
907 }
908 circle mincircle()//点集最小圆覆盖
909 {
910 random_shuffle(p,p+n);
911 point tri[4];
912 circle c;
913 solve(n,0,tri,c);
914 return c;
915 }
916 int circlecover(double r)//单位圆覆盖
917 {
918 int ans=0,i,j;
919 vector<pair<double,int> >v;
920 for (i=0;i<n;i++)
921 {
922 v.clear();
923 for (j=0;j<n;j++)if (i!=j)
924 {
925 point q=p[i].sub(p[j]);
926 double d=q.len();
927 if (dblcmp(d-2*r)<=0)
928 {
929 double arg=atan2(q.y,q.x);
930 if (dblcmp(arg)<0)arg+=2*pi;
931 double t=acos(d/(2*r));
932 v.push_back(make_pair(arg-t+2*pi,-1));
933 v.push_back(make_pair(arg+t+2*pi,1));
934 }
935 }
936 sort(v.begin(),v.end());
937 int cur=0;
938 for (j=0;j<v.size();j++)
939 {
940 if (v[j].second==-1)++cur;
941 else --cur;
942 ans=max(ans,cur);
943 }
944 }
945 return ans+1;
946 }
947 int pointinpolygon(point q)//点在凸多边形内部的判定
948 {
949 if (getdir())reverse(p,p+n);
950 if (dblcmp(q.sub(p[0]).det(p[n-1].sub(p[0])))==0)
951 {
952 if (line(p[n-1],p[0]).pointonseg(q))return n-1;
953 return -1;
954 }
955 int low=1,high=n-2,mid;
956 while (low<=high)
957 {
958 mid=(low+high)>>1;
959 if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>=0&&dblcmp(q.sub(p[0]).det(p[mid+1].sub(p[0])))<0)
960 {
961 polygon c;
962 c.p[0]=p[mid];
963 c.p[1]=p[mid+1];
964 c.p[2]=p[0];
965 c.n=3;
966 if (c.relationpoint(q))return mid;
967 return -1;
968 }
969 if (dblcmp(q.sub(p[0]).det(p[mid].sub(p[0])))>0)
970 {
971 low=mid+1;
972 }
973 else
974 {
975 high=mid-1;
976 }
977 }
978 return -1;
979 }
980
981 //ADD
982 //最小矩形面积覆盖
983 //A必须是凸包（而且是逆时针顺序）
984 //Uva 10173
985 double cross(point A,point B,point C)
986 {}
987 double dot(point A,point B,point C)
988 {}
989 double minRectangleCover(polygon A)
990 {}
991
992 //ADD
993 //直线切凸多边形
994 //多边形是逆时针的，在q1q2的左侧
995 //HDU3982
996 /*
997 vector<point> convexcut(const vector<point> &ps,point q1,point q2)
998 {
999 vector<point> qs;
1000 int n=ps.size();
1001 for (int i=0;i<n;i++)
1002 {
1003 point p1=ps[i],p2=ps[(i+1)%n];
1004 int d1=sgn((q2-q1)^(p1-q1)),d2=sgn((q2-q1)^(p2-q1));
1005 if (d1>=0)
1006 qs.push_back(p1);
1007 if (d1*d2<0)
1008 qs.push_back(line(p1,p2).crosspoint(line(q1,q2)));
1009 }
1010 return qs;
1011 }
1012 */
1013 };
1014
1015 //ADD
1016 //直线切凸多边形
1017 //多边形是逆时针的，在q1q2的左侧
1018 //HDU3982
1019 vector<point> convexcut(const vector<point> &ps,point q1,point q2)
1020 {
1021 vector<point> qs;
1022 int n=ps.size();
1023 for (int i=0; i<n; i++)
1024 {
1025 point p1=ps[i],p2=ps[(i+1)%n];
1026 int d1=sgn((q2-q1)^(p1-q1)),d2=sgn((q2-q1)^(p2-q1));
1027 if (d1>=0)
1028 qs.push_back(p1);
1029 if (d1*d2<0)
1030 qs.push_back(line(p1,p2).crosspoint(line(q1,q2)));
1031 }
1032 return qs;
1033 }
1034
1035 double CutLine[10000][10];
1036 point Cherry;
1037 int TotalTimes,n;
1038 double r;
1039
1040 int main()
1041 {
1042 freopen("in.txt","r",stdin);
1043
1044 cin>>TotalTimes;
1045 for (int Times=1;Times<=TotalTimes;Times++)
1046 {
1047 //cin>>r>>n;
1048 scanf("%lf%d",&r,&n);
1049 circle Cake=circle(point(0.00,0.00),r);
1050 vector<point> BigPolygon;
1051 BigPolygon.push_back(point(-r,r));
1052 BigPolygon.push_back(point(-r,-r));
1053 BigPolygon.push_back(point(r,-r));
1054 BigPolygon.push_back(point(r,r));
1055
1056 for (int i=1;i<=n;i++)
1057 scanf("%lf%lf%lf%lf",&CutLine[i][1],&CutLine[i][2],&CutLine[i][3],&CutLine[i][4]);
1058 //cin>>CutLine[i][1]>>CutLine[i][2]>>CutLine[i][3]>>CutLine[i][4];
1059
1060 Cherry.input();
1061
1062 for (int i=1;i<=n;i++)
1063 {
1064 line cut(point(CutLine[i][1],CutLine[i][2]),point(CutLine[i][3],CutLine[i][4]));
1065 if (cut.relation(Cherry)==2)
1066 {
1067 //cut=line(point(CutLine[i][3],CutLine[i][4]),point(CutLine[i][1],CutLine[i][2]));
1068 BigPolygon=convexcut(BigPolygon,point(CutLine[i][3],CutLine[i][4]),point(CutLine[i][1],CutLine[i][2]));
1069 }
1070 else
1071 {
1072 BigPolygon=convexcut(BigPolygon,point(CutLine[i][1],CutLine[i][2]),point(CutLine[i][3],CutLine[i][4]));
1073 }
1074 }
1075
1076 polygon Bigpolygon; Bigpolygon.n=0;
1077 for (vector<point>::iterator i=BigPolygon.begin();i!=BigPolygon.end();i++)
1078 {
1079 point tmp=*i;
1080 Bigpolygon.add(tmp);
1081 }
1082 //double ans=Bigpolygon.getarea();
1083 double CakeArea=Cake.area();
1084 double ans=Bigpolygon.areacircle(Cake);
1085 ans=ans/CakeArea*100;
1086 printf("Case %d: %.5lf%%\n",Times,ans);
1087 }
1088 return 0;
1089 }
Reference：http://blog.csdn.net/zxy_snow/article/details/6739561
话说自从开始刷计算几何之后发现自己代码风格越来越屎了，满满的工程代码既视感。。【逃
时间： 2024-10-22 06:12:37