题意：给出一个闭合折线上的一堆点（不按顺序），然后再给一个点P，要求判断P是否在闭合折线内
sol attempt1：一开始觉得是个模板题的，后来发现不对劲：
给出的点并不按照顺序。这样模板大法就不行了（geiline函数是按顺序建line的，会错乱掉）
sol attempt2：手艹大法：先建立好图形，然后再判断：
从点P向外引射线，若射线与图形的边相交奇数次，说明P在图形内。
PS：本题测试数据好像有点不对劲：
对于这组数据：
12 0 40 0 50 0 0 10 10 0 30 0 20 0 10 10 40 20 50 20 0 10 20 10 30 10 1
很容易画出图：
很容易发现绿色点P应该是INSIDE的，但是testdata是OUTSIDE
围观了下标程。。。好像是对于多组数据标程忘了初始化数组了。当时现场也并没有人过orz
附我的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-8;
32 const double pi=acos(-1.0);
33 const double inf=1e20;
34 const int maxp=10010; //多边形内点的数量，注意按需求调整
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 struct point
51 {
52 double x,y;
53 point(){}
54 point(double _x,double _y):
55 x(_x),y(_y){};
56 void input()
57 {
58 scanf("%lf%lf",&x,&y);
59 }
60 void output()
61 {
62 printf("%.2f %.2f\n",x,y);
63 }
64 bool operator==(point a)const
65 {
66 return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
67 }
68 bool operator<(point a)const
69 {
70 return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
71 }
72
73 point operator +(const point &b)const
74 {
75 return point(x+b.x,y+b.y);
76 }
77 point operator -(const point &b)const
78 {
79 return point(x-b.x,y-b.y);
80 }
81 point operator *(const double &k)const
82 {
83 return point(x*k,y*k);
84 }
85 point operator /(const double &k)const
86 {
87 return point(x/k,y/k);
88 }
89 double operator ^(const point &b)const
90 {
91 return x*b.y-y*b.x;
92 }
93 double operator *(const point &b)const
94 {
95 return x*b.x+y*b.y;
96 }
97
98 double len()
99 {
100 return hypot(x,y);
101 }
102 double len2()
103 {
104 return x*x+y*y;
105 }
106 double distance(point p)
107 {
108 return hypot(x-p.x,y-p.y);
109 }
110 point add(point p)
111 {
112 return point(x+p.x,y+p.y);
113 }
114 point sub(point p)
115 {
116 return point(x-p.x,y-p.y);
117 }
118 point mul(double b)
119 {
120 return point(x*b,y*b);
121 }
122 point div(double b)
123 {
124 return point(x/b,y/b);
125 }
126 double dot(point p)
127 {
128 return x*p.x+y*p.y;
129 }
130 double det(point p)
131 {
132 return x*p.y-y*p.x;
133 }
134 double rad(point a,point b)
135 {
136 point p=*this;
137 return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
138 }
139 point trunc(double r)
140 {
141 double l=len();
142 if (!dblcmp(l))return *this;
143 r/=l;
144 return point(x*r,y*r);
145 }
146 point rotleft()
147 {
148 return point(-y,x);
149 }
150 point rotright()
151 {
152 return point(y,-x);
153 }
154 point rotate(point p,double angle)//绕点p逆时针旋转angle角度
155 {
156 point v=this->sub(p);
157 double c=cos(angle),s=sin(angle);
158 return point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
159 }
160 };
161
162 struct line
163 {
164 point a,b;
165 line(){}
166 line(point _a,point _b)
167 {
168 a=_a;
169 b=_b;
170 }
171 bool operator==(line v)
172 {
173 return (a==v.a)&&(b==v.b);
174 }
175 //倾斜角angle
176 line(point p,double angle)
177 {
178 a=p;
179 if (dblcmp(angle-pi/2)==0)
180 {
181 b=a.add(point(0,1));
182 }
183 else
184 {
185 b=a.add(point(1,tan(angle)));
186 }
187 }
188 //ax+by+c=0
189 line(double _a,double _b,double _c)
190 {
191 if (dblcmp(_a)==0)
192 {
193 a=point(0,-_c/_b);
194 b=point(1,-_c/_b);
195 }
196 else if (dblcmp(_b)==0)
197 {
198 a=point(-_c/_a,0);
199 b=point(-_c/_a,1);
200 }
201 else
202 {
203 a=point(0,-_c/_b);
204 b=point(1,(-_c-_a)/_b);
205 }
206 }
207 void input()
208 {
209 a.input();
210 b.input();
211 }
212 void adjust()
213 {
214 if (b<a)swap(a,b);
215 }
216 double length()
217 {
218 return a.distance(b);
219 }
220 double angle()//直线倾斜角 0<=angle<180
221 {
222 double k=atan2(b.y-a.y,b.x-a.x);
223 if (dblcmp(k)<0)k+=pi;
224 if (dblcmp(k-pi)==0)k-=pi;
225 return k;
226 }
227 //点和线段关系
228 //1 在逆时针
229 //2 在顺时针
230 //3 平行
231 int relation(point p)
232 {
233 int c=dblcmp(p.sub(a).det(b.sub(a)));
234 if (c<0)return 1;
235 if (c>0)return 2;
236 return 3;
237 }
238 bool pointonseg(point p)
239 {
240 return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
241 }
242 bool parallel(line v)
243 {
244 return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
245 }
246 //2 规范相交
247 //1 非规范相交
248 //0 不相交
249 int segcrossseg(line v)
250 {
251 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
252 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
253 int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
254 int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
255 if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
256 return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
257 d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
258 d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
259 d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
260 }
261 int linecrossseg(line v)//*this seg v line
262 {
263 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
264 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
265 if ((d1^d2)==-2)return 2;
266 return (d1==0||d2==0);
267 }
268 //0 平行
269 //1 重合
270 //2 相交
271 int linecrossline(line v)
272 {
273 if ((*this).parallel(v))
274 {
275 return v.relation(a)==3;
276 }
277 return 2;
278 }
279 point crosspoint(line v)
280 {
281 double a1=v.b.sub(v.a).det(a.sub(v.a));
282 double a2=v.b.sub(v.a).det(b.sub(v.a));
283 return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
284 }
285 double dispointtoline(point p)
286 {
287 return fabs(p.sub(a).det(b.sub(a)))/length();
288 }
289 double dispointtoseg(point p)
290 {
291 if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
292 {
293 return min(p.distance(a),p.distance(b));
294 }
295 return dispointtoline(p);
296 }
297 point lineprog(point p)
298 {
299 return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a))/b.sub(a).len2()));
300 }
301 point symmetrypoint(point p)
302 {
303 point q=lineprog(p);
304 return point(2*q.x-p.x,2*q.y-p.y);
305 }
306 };
307
308 struct Vector:public point
309 {
310 Vector(){}
311 Vector(double a,double b)
312 {
313 x=a; y=b;
314 }
315 Vector(point _a,point _b) //a->b
316 {
317 double dx=_b.x-_a.x;
318 double dy=_b.y-_a.y;
319 x=dx; y=dy;
320 }
321 Vector(line v)
322 {
323 double dx=v.b.x-v.a.x;
324 double dy=v.b.y-v.a.y;
325 x=dx; y=dy;
326 }
327 double length()
328 {
329 return (sqrt(x*x+y*y));
330 }
331 Vector Normal()
332 {
333 double L=sqrt(x*x+y*y);
334 Vector Vans=Vector(-y/L,x/L);
335 return Vans;
336 }
337 };
338
339 struct polygon
340 {
341 int n; //the number of points
342 int nl; //the number of lines
343 point p[maxp]; //0..n-1
344 line l[maxp]; //0..nl-1
345 /*
346 void input(int X)
347 {
348 n=X;
349 nl=X;
350 for (int i=0;i<n;i++)
351 {
352 p[i].input();
353 }
354 }
355 */
356 void add(point q)
357 {
358 p[n++]=q;
359 }
360 /*
361 void getline()
362 {
363 for (int i=0;i<n;i++)
364 {
365 l[i]=line(p[i],p[(i+1)%n]);
366 }
367 }
368 */
369
370 //3 点上
371 //2 边上
372 //1 内部
373 //0 外部
374 int relationpoint(point q)
375 {
376 int i,j;
377 for (i=0;i<n;i++)
378 {
379 if (p[i]==q)return 3;
380 }
381 //getline();
382 for (i=0;i<nl;i++)
383 {
384 if (l[i].pointonseg(q))return 2;
385 }
386 int cnt=0;
387 for (i=0;i<n;i++)
388 {
389 j=(i+1)%n;
390 int k=dblcmp(q.sub(p[j]).det(p[i].sub(p[j])));
391 int u=dblcmp(p[i].y-q.y);
392 int v=dblcmp(p[j].y-q.y);
393 if (k>0&&u<0&&v>=0)cnt++;
394 if (k<0&&v<0&&u>=0)cnt--;
395 }
396 return cnt!=0;
397 }
398 /*
399 //1 在多边形内长度为正
400 //2 相交或与边平行
401 //0 无任何交点
402 int relationline(line u)
403 {
404 int i,j,k=0;
405 //getline();
406 for (i=0;i<nl;i++)
407 {
408 if (l[i].segcrossseg(u)==2)return 1;
409 if (l[i].segcrossseg(u)==1)k=1;
410 }
411 if (!k)return 0;
412 vector<point>vp;
413 for (i=0;i<nl;i++)
414 {
415 if (l[i].segcrossseg(u))
416 {
417 if (l[i].parallel(u))
418 {
419 vp.pb(u.a);
420 vp.pb(u.b);
421 vp.pb(l[i].a);
422 vp.pb(l[i].b);
423 continue;
424 }
425 vp.pb(l[i].crosspoint(u));
426 }
427 }
428 sort(vp.begin(),vp.end());
429 int sz=vp.size();
430 for (i=0;i<sz-1;i++)
431 {
432 point mid=vp[i].add(vp[i+1]).div(2);
433 if (relationpoint(mid)==1)return 1;
434 }
435 return 2;
436 */
437 };
438
439 int k;
440 double X0,Y0;
441 line Line[10010];
442 struct poi
443 {
444 int x; int y;
445 }a[10010];
446
447 void xsort(int l,int r)
448 {
449 int i=l,j=r;
450 int mid=(l+r)/2;
451 while(i<j)
452 {
453 while((a[i].x<a[mid].x)||((a[i].x==a[mid].x)&&(a[i].y<a[mid].y)))
454 i++;
455 while((a[j].x>a[mid].x)||((a[j].x==a[mid].x)&&(a[j].y>a[mid].y)))
456 j--;
457 if (i<=j)
458 {
459 swap(a[i].x,a[j].x);
460 swap(a[i].y,a[j].y);
461 i++; j--;
462 }
463 }
464 if (l<j) xsort(l,j);
465 if (i<r) xsort(i,r);
466 }
467 void ysort(int l,int r)
468 {
469 int i=l,j=r;
470 int mid=(l+r)/2;
471 while(i<j)
472 {
473 while((a[i].y<a[mid].y)||((a[i].y==a[mid].y)&&(a[i].x<a[mid].x)))
474 i++;
475 while((a[j].y>a[mid].y)||((a[j].y==a[mid].y)&&(a[j].x>a[mid].x)))
476 j--;
477 if (i<=j)
478 {
479 swap(a[i].x,a[j].x);
480 swap(a[i].y,a[j].y);
481 i++; j--;
482 }
483 }
484 if (l<j) ysort(l,j);
485 if (i<r) ysort(i,r);
486 }
487
488 bool cross(point a, point b, point c) {
489 //if ( ((a.y > c.y && b.y <= c.y) || (a.y <= c.y && b.y > c.y)) && c.x < a.x) {
490 if ( ( ( (dblcmp(a.y-c.y)==1) && (dblcmp(b.y-c.y)!=1) )||( (dblcmp(a.y-c.y)!=1) && (dblcmp(b.y-c.y)==1) ) ) && (dblcmp(c.x-a.x)==-1) ) {
491 return true;
492 }
493 return false;
494 }
495
496 int main()
497 {
498 while(cin>>k)
499 {
500 for (int i=1;i<=k;i++)
501 cin>>a[i].x>>a[i].y;
502
503 xsort(1,k);
504 int NL=0;
505 for (int i=1;i<=k-1;i=i+2)
506 {
507 NL++;
508 Line[NL]=line(point(a[i].x,a[i].y),point(a[i+1].x,a[i+1].y));
509 //cout<<a[i].x<<","<<a[i].y<<" "<<a[i+1].x<<","<<a[i+1].y<<endl;
510 }
511 ysort(1,k);
512 for (int i=1;i<=k-1;i=i+2)
513 {
514 NL++;
515 Line[NL]=line(point(a[i].x,a[i].y),point(a[i+1].x,a[i+1].y));
516 //cout<<a[i].x<<","<<a[i].y<<" "<<a[i+1].x<<","<<a[i+1].y<<endl;
517 }
518
519 cin>>X0>>Y0;
520 point des=point(X0,Y0);
521 int ans=99,cnt=0;
522 for (int i=1;i<=NL;i++)
523 {
524 line tl=Line[i];
525 if (tl.pointonseg(des))
526 {
527 //cout<<i<<endl;
528 ans=2;
529 }
530 if (cross(tl.a,tl.b,des))
531 {
532 cout<<i<<endl;
533 cnt++;
534 }
535 }
536 if (ans==2)
537 cout<<"BORDER"<<endl;
538 else
539 {
540 if(cnt%2==0)
541 cout<<"OUTSIDE"<<endl;
542 else
543 cout<<"INSIDE"<<endl;
544 }
545
546 /*
547 PY.n=k; PY.nl=NL;
548 for (int i=0;i<k;i++)
549 {
550 cout<<a[i+1].x<<","<<a[i+1].y<<endl;
551 PY.p[i]=point(a[i+1].x,a[i+1].y);
552 }
553 for (int i=0;i<NL;i++)
554 {
555 cout<<Line[i+1].a.x<<","<<Line[i+1].a.y<<"->"<<Line[i+1].b.x<<","<<Line[i+1].b.y<<endl;
556 PY.l[i]=Line[i+1];
557 }
558
559 cin>>X0>>Y0;
560 int ans=PY.relationpoint(point(X0,Y0));
561 //3 点上
562 //2 边上
563 //1 内部
564 //0 外部
565 if (ans==1)
566 cout<<"INSIDE"<<endl;
567 else if (ans==0)
568 cout<<"OUTSIDE"<<endl;
569 else
570 cout<<"BORDER"<<endl;
571 */
572 }
573 return 0;
574 }
时间： 2024-10-06 18:38:50