题意：给出一多边形。判断多边形是否存在一点，使得多边形边界上的所有点都能看见该点。
sol：在纸上随手画画就可以找出规律：按逆时针顺序连接所有点。然后找出这些line的半平面交。
题中给出的点已经按顺时针排好序了，所以只要倒过来一下就可以了。很简单的模板题。
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=1111;
35 int dblcmp(double d)
36 {
37 if (fabs(d)<eps)return 0;
38 return d>eps?1:-1;
39 }
40 inline double sqr(double x){return x*x;}
41 struct point
42 {
43 double x,y;
44 point(){}
45 point(double _x,double _y):
46 x(_x),y(_y){};
47 void input()
48 {
49 scanf("%lf%lf",&x,&y);
50 }
51 void output()
52 {
53 printf("%.2f %.2f\n",x,y);
54 }
55 bool operator==(point a)const
56 {
57 return dblcmp(a.x-x)==0&&dblcmp(a.y-y)==0;
58 }
59 bool operator<(point a)const
60 {
61 return dblcmp(a.x-x)==0?dblcmp(y-a.y)<0:x<a.x;
62 }
63 double len()
64 {
65 return hypot(x,y);
66 }
67 double len2()
68 {
69 return x*x+y*y;
70 }
71 double distance(point p)
72 {
73 return hypot(x-p.x,y-p.y);
74 }
75 point add(point p)
76 {
77 return point(x+p.x,y+p.y);
78 }
79 point sub(point p)
80 {
81 return point(x-p.x,y-p.y);
82 }
83 point mul(double b)
84 {
85 return point(x*b,y*b);
86 }
87 point div(double b)
88 {
89 return point(x/b,y/b);
90 }
91 double dot(point p)
92 {
93 return x*p.x+y*p.y;
94 }
95 double det(point p)
96 {
97 return x*p.y-y*p.x;
98 }
99 double rad(point a,point b)
100 {
101 point p=*this;
102 return fabs(atan2(fabs(a.sub(p).det(b.sub(p))),a.sub(p).dot(b.sub(p))));
103 }
104 point trunc(double r)
105 {
106 double l=len();
107 if (!dblcmp(l))return *this;
108 r/=l;
109 return point(x*r,y*r);
110 }
111 point rotleft()
112 {
113 return point(-y,x);
114 }
115 point rotright()
116 {
117 return point(y,-x);
118 }
119 point rotate(point p,double angle)//绕点p逆时针旋转angle角度
120 {
121 point v=this->sub(p);
122 double c=cos(angle),s=sin(angle);
123 return point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
124 }
125 };
126 struct line
127 {
128 point a,b;
129 line(){}
130 line(point _a,point _b)
131 {
132 a=_a;
133 b=_b;
134 }
135 bool operator==(line v)
136 {
137 return (a==v.a)&&(b==v.b);
138 }
139 //倾斜角angle
140 line(point p,double angle)
141 {
142 a=p;
143 if (dblcmp(angle-pi/2)==0)
144 {
145 b=a.add(point(0,1));
146 }
147 else
148 {
149 b=a.add(point(1,tan(angle)));
150 }
151 }
152 //ax+by+c=0
153 line(double _a,double _b,double _c)
154 {
155 if (dblcmp(_a)==0)
156 {
157 a=point(0,-_c/_b);
158 b=point(1,-_c/_b);
159 }
160 else if (dblcmp(_b)==0)
161 {
162 a=point(-_c/_a,0);
163 b=point(-_c/_a,1);
164 }
165 else
166 {
167 a=point(0,-_c/_b);
168 b=point(1,(-_c-_a)/_b);
169 }
170 }
171 void input()
172 {
173 a.input();
174 b.input();
175 }
176 void adjust()
177 {
178 if (b<a)swap(a,b);
179 }
180 double length()
181 {
182 return a.distance(b);
183 }
184 double angle()//直线倾斜角 0<=angle<180
185 {
186 double k=atan2(b.y-a.y,b.x-a.x);
187 if (dblcmp(k)<0)k+=pi;
188 if (dblcmp(k-pi)==0)k-=pi;
189 return k;
190 }
191 //点和线段关系
192 //1 在逆时针
193 //2 在顺时针
194 //3 平行
195 int relation(point p)
196 {
197 int c=dblcmp(p.sub(a).det(b.sub(a)));
198 if (c<0)return 1;
199 if (c>0)return 2;
200 return 3;
201 }
202 bool pointonseg(point p)
203 {
204 return dblcmp(p.sub(a).det(b.sub(a)))==0&&dblcmp(p.sub(a).dot(p.sub(b)))<=0;
205 }
206 bool parallel(line v)
207 {
208 return dblcmp(b.sub(a).det(v.b.sub(v.a)))==0;
209 }
210 //2 规范相交
211 //1 非规范相交
212 //0 不相交
213 int segcrossseg(line v)
214 {
215 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
216 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
217 int d3=dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
218 int d4=dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
219 if ((d1^d2)==-2&&(d3^d4)==-2)return 2;
220 return (d1==0&&dblcmp(v.a.sub(a).dot(v.a.sub(b)))<=0||
221 d2==0&&dblcmp(v.b.sub(a).dot(v.b.sub(b)))<=0||
222 d3==0&&dblcmp(a.sub(v.a).dot(a.sub(v.b)))<=0||
223 d4==0&&dblcmp(b.sub(v.a).dot(b.sub(v.b)))<=0);
224 }
225 int linecrossseg(line v)//*this seg v line
226 {
227 int d1=dblcmp(b.sub(a).det(v.a.sub(a)));
228 int d2=dblcmp(b.sub(a).det(v.b.sub(a)));
229 if ((d1^d2)==-2)return 2;
230 return (d1==0||d2==0);
231 }
232 //0 平行
233 //1 重合
234 //2 相交
235 int linecrossline(line v)
236 {
237 if ((*this).parallel(v))
238 {
239 return v.relation(a)==3;
240 }
241 return 2;
242 }
243 point crosspoint(line v)
244 {
245 double a1=v.b.sub(v.a).det(a.sub(v.a));
246 double a2=v.b.sub(v.a).det(b.sub(v.a));
247 return point((a.x*a2-b.x*a1)/(a2-a1),(a.y*a2-b.y*a1)/(a2-a1));
248 }
249 double dispointtoline(point p)
250 {
251 return fabs(p.sub(a).det(b.sub(a)))/length();
252 }
253 double dispointtoseg(point p)
254 {
255 if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
256 {
257 return min(p.distance(a),p.distance(b));
258 }
259 return dispointtoline(p);
260 }
261 point lineprog(point p)
262 {
263 return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a))/b.sub(a).len2()));
264 }
265 point symmetrypoint(point p)
266 {
267 point q=lineprog(p);
268 return point(2*q.x-p.x,2*q.y-p.y);
269 }
270 };
271
272 struct Vector:public point
273 {
274 Vector(){}
275 Vector(double a,double b)
276 {
277 x=a; y=b;
278 }
279 Vector(point _a,point _b) //a->b
280 {
281 double dx=_b.x-_a.x;
282 double dy=_b.y-_a.y;
283 x=dx; y=dy;
284 }
285 Vector(line v)
286 {
287 double dx=v.b.x-v.a.x;
288 double dy=v.b.y-v.a.y;
289 x=dx; y=dy;
290 }
291 double length()
292 {
293 return (sqrt(x*x+y*y));
294 }
295 Vector Normal()
296 {
297 double L=sqrt(x*x+y*y);
298 Vector Vans=Vector(-y/L,x/L);
299 return Vans;
300 }
301 };
302
303 struct halfplane:public line //半平面
304 {
305 double angle;
306 halfplane(){}
307 //表示向量 a->b逆时针(左侧)的半平面
308 halfplane(point _a,point _b)
309 {
310 a=_a;
311 b=_b;
312 }
313 halfplane(line v)
314 {
315 a=v.a;
316 b=v.b;
317 }
318 void calcangle()
319 {
320 angle=atan2(b.y-a.y,b.x-a.x);
321 }
322 bool operator<(const halfplane &b)const
323 {
324 return angle<b.angle;
325 }
326 };
327 struct halfplanes //半平面交
328 {
329 int n;
330 halfplane hp[maxp];
331 point p[maxp];
332 int que[maxp];
333 int st,ed;
334 void push(halfplane tmp)
335 {
336 hp[n++]=tmp;
337 }
338 void unique()
339 {
340 int m=1,i;
341 for (i=1;i<n;i++)
342 {
343 if (dblcmp(hp[i].angle-hp[i-1].angle))hp[m++]=hp[i];
344 else if (dblcmp(hp[m-1].b.sub(hp[m-1].a).det(hp[i].a.sub(hp[m-1].a))>0))hp[m-1]=hp[i];
345 }
346 n=m;
347 }
348 bool halfplaneinsert()
349 {
350 int i;
351 for (i=0;i<n;i++)hp[i].calcangle();
352 sort(hp,hp+n);
353 unique();
354 que[st=0]=0;
355 que[ed=1]=1;
356 p[1]=hp[0].crosspoint(hp[1]);
357 for (i=2;i<n;i++)
358 {
359 while (st<ed&&dblcmp((hp[i].b.sub(hp[i].a).det(p[ed].sub(hp[i].a))))<0)ed--;
360 while (st<ed&&dblcmp((hp[i].b.sub(hp[i].a).det(p[st+1].sub(hp[i].a))))<0)st++;
361 que[++ed]=i;
362 if (hp[i].parallel(hp[que[ed-1]]))return false;
363 p[ed]=hp[i].crosspoint(hp[que[ed-1]]);
364 }
365 while (st<ed&&dblcmp(hp[que[st]].b.sub(hp[que[st]].a).det(p[ed].sub(hp[que[st]].a)))<0)ed--;
366 while (st<ed&&dblcmp(hp[que[ed]].b.sub(hp[que[ed]].a).det(p[st+1].sub(hp[que[ed]].a)))<0)st++;
367 if (st+1>=ed)return false;
368 return true;
369 }
370 /*
371 void getconvex(polygon &con)
372 {
373 p[st]=hp[que[st]].crosspoint(hp[que[ed]]);
374 con.n=ed-st+1;
375 int j=st,i=0;
376 for (;j<=ed;i++,j++)
377 {
378 con.p[i]=p[j];
379 }
380 }*/
381 };
382
383 point p[1000];
384 halfplanes TH;
385 int n,T;
386
387 int main()
388 {
389 //freopen("in.txt","r",stdin);
390
391 cin>>T;
392 while (T--)
393 {
394 cin>>n;
395 for (int i=n-1;i>=0;i--)
396 p[i].input();
397 //p[i]->p[i+1]
398
399 TH.n=0;
400 for (int i=0;i<=n-1;i++)
401 TH.push(halfplane(p[i],p[(i+1)%n]));
402
403 if (TH.halfplaneinsert())
404 cout<<"YES"<<endl;
405 else cout<<"NO"<<endl;
406 }
407
408 return 0;
409 }
时间： 2024-10-12 23:49:48