poj 3335 /poj 3130/ poj 1474 半平面交 判断核是否存在 / poj1279 半平面交 求核的面积


  1 /***************

  2 poj 3335  点序顺时针

  3 ***************/

  4 #include <iostream>

  5 #include <cmath>

  6 #include <algorithm>

  7 using namespace std;

  8 const double eps = 1e-8;

  9 const double maxn = 0x7f7f7f7f;

 10 int dcmp(double x){

 11     if(fabs(x)<eps)

 12         return 0;

 13     else

 14         return x<0?-1:1;

 15 }

 16 struct point {

 17     double x,y;

 18     point (double x=0,double y =0):x(x),y(y){}

 19 };

 20 point p[150];

 21 typedef point Vector;

 22

 23 struct polygon{

 24     point p[150];

 25     int Size;

 26 };

 27

 28 struct line{

 29     point fir,sec;

 30     line(point a = point(),point b = point()){

 31         fir = a;

 32         sec = b;

 33     }

 34 };

 35

 36 Vector operator -(point a,point b){

 37     return Vector(a.x-b.x,a.y-b.y);

 38 }

 39

 40 Vector operator *(Vector a,double p){

 41     return Vector (a.x*p,a.y*p);

 42 }

 43 Vector operator + (Vector a,Vector b){

 44     return Vector (a.x+b.x,a.y+b.y);

 45 }

 46

 47 double cross(Vector a,Vector b){

 48     return a.x*b.y-a.y*b.x;

 49 }

 50

 51 double dot(Vector a,Vector b){

 52     return a.x*b.x+a.y*b.y;

 53 }

 54

 55 point getLineIntersection(point p,Vector v,point q,Vector w){

 56     Vector u = p-q;

 57     double t = cross(w,u)/cross(v,w);

 58     return p+v*t;

 59 }

 60

 61 bool onsegment(point p,point a1,point a2){

 62     return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;

 63 }

 64

 65 polygon cutploygon(polygon poly,line ln){

 66     polygon newploy;

 67     int m=0;

 68     int n = poly.Size;

 69     point a = ln.fir,b = ln.sec;

 70     for(int i=0;i<n;i++){

 71         point c = poly.p[i];

 72         point d = poly.p[(i+1)%n];

 73         double cc = cross(b-a,c-a);

 74         double dd = cross(b-a,d-a);

 75         if(cc>=0)

 76             newploy.p[m++] = c;

 77         if(cc*dd<0)

 78             newploy.p[m++] = getLineIntersection(a,b-a,c,d-c);

 79     }

 80     newploy.Size = m;

 81     return newploy;

 82 }

 83

 84 int main()

 85 {

 86     int t;

 87     cin>>t;

 88     int n;

 89     while(t--){

 90         cin>>n;

 91         for(int i=0;i<n;i++)

 92             cin>>p[i].x>>p[i].y;

 93         polygon poly;

 94         poly.Size = 4;

 95         poly.p[0].x = -maxn;

 96         poly.p[0].y = -maxn;

 97         poly.p[1].x = maxn;

 98         poly.p[1].y = -maxn;

 99         poly.p[2].x = maxn;

100         poly.p[2].y = maxn;

101         poly.p[3].x = -maxn;

102         poly.p[3].y = maxn;

103         bool flag = true;

104         for(int i=1;i<=n;i++){

105             line ln;

106             ln.fir = p[i%n];

107             ln.sec = p[i-1];

108             poly = cutploygon(poly,ln);

109             if(poly.Size==0){

110                 flag = false;

111                 break;

112             }

113         }

114         if(!flag)

115             cout<<"NO"<<endl;

116         else

117             cout<<"YES"<<endl;

118     }

119     return 0;

120 }

121

122 /****************************************/

123 poj 3130  点序逆时针

124 /****************************************/

125

126 #include <iostream>

127 #include <cmath>

128 #include <algorithm>

129 using namespace std;

130 const double eps = 1e-8;

131 const double maxn = 0x7f7f7f7f;

132 int dcmp(double x){

133     if(fabs(x)<eps)

134         return 0;

135     else

136         return x<0?-1:1;

137 }

138 struct point {

139     double x,y;

140     point (double x=0,double y =0):x(x),y(y){}

141 };

142 point p[150];

143 typedef point Vector;

144

145 struct polygon{

146     point p[150];

147     int Size;

148 };

149

150 struct line{

151     point fir,sec;

152     line(point a = point(),point b = point()){

153         fir = a;

154         sec = b;

155     }

156 };

157

158 Vector operator -(point a,point b){

159     return Vector(a.x-b.x,a.y-b.y);

160 }

161

162 Vector operator *(Vector a,double p){

163     return Vector (a.x*p,a.y*p);

164 }

165 Vector operator + (Vector a,Vector b){

166     return Vector (a.x+b.x,a.y+b.y);

167 }

168

169 double cross(Vector a,Vector b){

170     return a.x*b.y-a.y*b.x;

171 }

172

173 double dot(Vector a,Vector b){

174     return a.x*b.x+a.y*b.y;

175 }

176

177 point getLineIntersection(point p,Vector v,point q,Vector w){

178     Vector u = p-q;

179     double t = cross(w,u)/cross(v,w);

180     return p+v*t;

181 }

182

183 bool onsegment(point p,point a1,point a2){

184     return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;

185 }

186

187 polygon cutploygon(polygon poly,line ln){

188     polygon newploy;

189     int m=0;

190     int n = poly.Size;

191     point a = ln.fir,b = ln.sec;

192     for(int i=0;i<n;i++){

193         point c = poly.p[i];

194         point d = poly.p[(i+1)%n];

195         double cc = cross(b-a,c-a);

196         double dd = cross(b-a,d-a);

197         if(cc>=0)

198             newploy.p[m++] = c;

199         if(cc*dd<0)

200             newploy.p[m++] = getLineIntersection(a,b-a,c,d-c);

201     }

202     newploy.Size = m;

203     return newploy;

204 }

205

206 int main()

207 {

208     int n;

209     while(cin>>n&&n){

210         for(int i=0;i<n;i++)

211             cin>>p[i].x>>p[i].y;

212         polygon poly;

213         poly.Size = 4;

214         poly.p[0].x = -maxn;

215         poly.p[0].y = -maxn;

216         poly.p[1].x = maxn;

217         poly.p[1].y = -maxn;

218         poly.p[2].x = maxn;

219         poly.p[2].y = maxn;

220         poly.p[3].x = -maxn;

221         poly.p[3].y = maxn;

222         bool flag = true;

223         for(int i=0;i<n;i++){

224             line ln;

225             ln.fir = p[i%n];

226             ln.sec = p[(i+1)%n];

227             poly = cutploygon(poly,ln);

228             if(poly.Size==0){

229                 flag = false;

230                 break;

231             }

232         }

233         if(!flag)

234             cout<<"0"<<endl;

235         else

236             cout<<"1"<<endl;

237     }

238     return 0;

239 }

240

241

242 /*************************************/

243 poj  1474  点序顺时针

244 /*************************************/

245 #include <iostream>

246 #include <cmath>

247 #include <algorithm>

248 using namespace std;

249 const double eps = 1e-8;

250 const double maxn = 0x7f7f7f7f;

251 int dcmp(double x){

252     if(fabs(x)<eps)

253         return 0;

254     else

255         return x<0?-1:1;

256 }

257 struct point {

258     double x,y;

259     point (double x=0,double y =0):x(x),y(y){}

260 };

261 point p[150];

262 typedef point Vector;

263

264 struct polygon{

265     point p[150];

266     int Size;

267 };

268

269 struct line{

270     point fir,sec;

271     line(point a = point(),point b = point()){

272         fir = a;

273         sec = b;

274     }

275 };

276

277 Vector operator -(point a,point b){

278     return Vector(a.x-b.x,a.y-b.y);

279 }

280

281 Vector operator *(Vector a,double p){

282     return Vector (a.x*p,a.y*p);

283 }

284 Vector operator + (Vector a,Vector b){

285     return Vector (a.x+b.x,a.y+b.y);

286 }

287

288 double cross(Vector a,Vector b){

289     return a.x*b.y-a.y*b.x;

290 }

291

292 double dot(Vector a,Vector b){

293     return a.x*b.x+a.y*b.y;

294 }

295

296 point getLineIntersection(point p,Vector v,point q,Vector w){

297     Vector u = p-q;

298     double t = cross(w,u)/cross(v,w);

299     return p+v*t;

300 }

301

302 bool onsegment(point p,point a1,point a2){

303     return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;

304 }

305

306 polygon cutploygon(polygon poly,line ln){

307     polygon newploy;

308     int m=0;

309     int n = poly.Size;

310     point a = ln.fir,b = ln.sec;

311     for(int i=0;i<n;i++){

312         point c = poly.p[i];

313         point d = poly.p[(i+1)%n];

314         double cc = cross(b-a,c-a);

315         double dd = cross(b-a,d-a);

316         if(cc>=0)

317             newploy.p[m++] = c;

318         if(cc*dd<0)

319             newploy.p[m++] = getLineIntersection(a,b-a,c,d-c);

320     }

321     newploy.Size = m;

322     return newploy;

323 }

324

325 int main()

326 {

327     int n;

328     int cnt =1;

329     while(cin>>n&&n){

330         for(int i=0;i<n;i++)

331             cin>>p[i].x>>p[i].y;

332         polygon poly;

333         poly.Size = 4;

334         poly.p[0].x = -maxn;

335         poly.p[0].y = -maxn;

336         poly.p[1].x = maxn;

337         poly.p[1].y = -maxn;

338         poly.p[2].x = maxn;

339         poly.p[2].y = maxn;

340         poly.p[3].x = -maxn;

341         poly.p[3].y = maxn;

342         bool flag = true;

343         for(int i=1;i<=n;i++){

344             line ln;

345             ln.fir = p[i%n];

346             ln.sec = p[i-1];

347             poly = cutploygon(poly,ln);

348             if(poly.Size==0){

349                 flag = false;

350                 break;

351             }

352         }

353         //cout<<poly.Size<<endl;

354         cout<<"Floor #"<<cnt++<<endl;

355         if(!flag)

356             cout<<"Surveillance is impossible."<<endl;

357         else

358             cout<<"Surveillance is possible."<<endl;

359         cout<<endl;

360     }

361     return 0;

362 }

363

364

365 /**********************************

366 poj 1279  点序顺时针

367 **********************************/

368 #include <iostream>

369 #include <cmath>

370 #include <algorithm>

371 #include <cstdio>

372 using namespace std;

373 const double eps = 1e-8;

374 const double maxn = 0x7f7f7f7f;

375 int dcmp(double x){

376     if(fabs(x)<eps)

377         return 0;

378     else

379         return x<0?-1:1;

380 }

381 struct point {

382     double x,y;

383     point (double x=0,double y =0):x(x),y(y){}

384 };

385 point p[2000];

386 typedef point Vector;

387

388 struct polygon{

389     point p[2000];

390     int Size;

391 };

392

393 struct line{

394     point fir,sec;

395     line(point a = point(),point b = point()){

396         fir = a;

397         sec = b;

398     }

399 };

400

401 Vector operator -(point a,point b){

402     return Vector(a.x-b.x,a.y-b.y);

403 }

404

405 Vector operator *(Vector a,double p){

406     return Vector (a.x*p,a.y*p);

407 }

408 Vector operator + (Vector a,Vector b){

409     return Vector (a.x+b.x,a.y+b.y);

410 }

411

412 double cross(Vector a,Vector b){

413     return a.x*b.y-a.y*b.x;

414 }

415

416 double dot(Vector a,Vector b){

417     return a.x*b.x+a.y*b.y;

418 }

419

420 point getLineIntersection(point p,Vector v,point q,Vector w){

421     Vector u = p-q;

422     double t = cross(w,u)/cross(v,w);

423     return p+v*t;

424 }

425

426 bool onsegment(point p,point a1,point a2){

427     return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;

428 }

429

430 polygon cutploygon(polygon poly,line ln){

431     polygon newploy;

432     int m=0;

433     int n = poly.Size;

434     point a = ln.fir,b = ln.sec;

435     for(int i=0;i<n;i++){

436         point c = poly.p[i];

437         point d = poly.p[(i+1)%n];

438         double cc = cross(b-a,c-a);

439         double dd = cross(b-a,d-a);

440         if(cc>=0)

441             newploy.p[m++] = c;

442         if(cc*dd<0)

443             newploy.p[m++] = getLineIntersection(a,b-a,c,d-c);

444     }

445     newploy.Size = m;

446     return newploy;

447 }

448

449 double polyArea(point *p,int n){

450     double area =0;

451     for(int i=1;i<n-1;i++){

452         area += cross(p[i]-p[0],p[i+1]-p[0]);

453     }

454     return area/2;

455 }

456

457 int main()

458 {

459     int t;

460     cin>>t;

461     int n;

462     while(t--){

463         cin>>n;

464         for(int i=0;i<n;i++)

465             cin>>p[i].x>>p[i].y;

466         polygon poly;

467         poly.Size = 4;

468         poly.p[0].x = -maxn;

469         poly.p[0].y = -maxn;

470         poly.p[1].x = maxn;

471         poly.p[1].y = -maxn;

472         poly.p[2].x = maxn;

473         poly.p[2].y = maxn;

474         poly.p[3].x = -maxn;

475         poly.p[3].y = maxn;

476         bool flag = true;

477         for(int i=1;i<=n;i++){

478             line ln;

479             ln.fir = p[i%n];

480             ln.sec = p[i-1];

481             poly = cutploygon(poly,ln);

482             if(poly.Size==0){

483                 flag = false;

484                 break;

485             }

486         }

487         double res;

488         if(!flag)

489             res =0;

490         else{

491             res = polyArea(poly.p,poly.Size);

492         }

493         printf("%.2lf\n",res);

494     }

495     return 0;

496 }

poj 3335 /poj 3130/ poj 1474 半平面交 判断核是否存在 / poj1279 半平面交
求核的面积

时间： 2024-10-11 21:52:51

poj 3335 /poj 3130/ poj 1474 半平面交 判断核是否存在 / poj1279 半平面交 求核的面积的相关文章

半平面交 模板 poj 3335 poj 3130 poj 1474 判断半平面交是否为空集

半平面交模板 const double pi= acos(-1.0); #define arc(x) (x / 180 * pi) const double EPS = 1e-8; const int Max_N = 105; struct Point{ double x,y; Point(){} Point(double x, double y):x(x),y(y){} Point operator - (Point p){ return Point(x- p.x , y - p.y ) ;

poj 3335 Rotating Scoreboard（半平面交）

Rotating Scoreboard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6420   Accepted: 2550 Description This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the ed

POJ 3335 半平面交求多边形的核

Rotating Scoreboard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4899   Accepted: 1946 Description This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the ed

POJ 3335

/*半平面交求核心的增量法: 假设前N-1个半平面交,对于第N个半平面,只需用它来交前N-1个平面交出的多边形. 算法开始时,调整点的方向为顺时针方向,对于是否为顺时针,只需求出其面积,若为正,必为逆时针的. 对于每相邻两点求出一条直线,用该直线去交其半平面,并求出交点及判断原多边形点的方位. */ #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #i

POJ 3304 Segments 基础线段交判断

LINK 题意:询问是否存在直线,使得所有线段在其上的投影拥有公共点 思路:如果投影拥有公共区域,那么从投影的公共区域作垂线,显然能够与所有线段相交,那么题目转换为询问是否存在直线与所有线段相交.判断相交先求叉积再用跨立定理.枚举每个线段的起始结束点作为直线起点终点遍历即可. /** @Date : 2017-07-12 14:35:44 * @FileName: POJ 3304 基础线段交判断.cpp * @Platform: Windows * @Author : Lweleth ([em

POJ 3177 Redundant Paths POJ 3352 Road Construction（双连通）

POJ 3177 Redundant Paths POJ 3352 Road Construction 题目链接 题意:两题一样的,一份代码能交,给定一个连通无向图,问加几条边能使得图变成一个双连通图 思路:先求双连通,缩点后,计算入度为1的个数,然后(个数 + 1) / 2 就是答案(这题由于是只有一个连通块所以可以这么搞,如果有多个,就不能这样搞了) 代码: #include <cstdio> #include <cstring> #include <algorithm&

POJ 2236 Wireless Network ||POJ 1703 Find them, Catch them 并查集

POJ 2236 Wireless Network http://poj.org/problem?id=2236 题目大意: 给你N台损坏的电脑坐标,这些电脑只能与不超过距离d的电脑通信,但如果x和y均能C通信,则x和y可以通信.现在给出若干个操作, O p 代表修复编号为p的电脑 S p q代表询问p和q是不是能通信. 思路: 并查集即可.. 如果修复了一台电脑,则把与它相连距离不超过d的且修复了的放在一个集合里面. #include<cstdio> #include<cstring&

hdu 3934&amp;&amp;poj 2079 （凸包+旋转卡壳+求最大三角形面积）

链接:http://poj.org/problem?id=2079 Triangle Time Limit: 3000MS   Memory Limit: 30000K Total Submissions: 8173   Accepted: 2423 Description Given n distinct points on a plane, your task is to find the triangle that have the maximum area, whose vertices

poj 1163 The Triangle &amp;poj 3167 Cow Bowling (dp)

链接:poj 1163 题意:输入一个n层的三角形,第i层有i个数,求从第1层到第n层的所有路线中,权值之和最大的路线. 规定:第i层的某个数只能连线走到第i+1层中与它位置相邻的两个数中的一个. 状态方程:f[i][j]=max(f[i-1][j-1],f[i-1][j])+a[i][j]; 1163代码: #include<stdio.h> #include<string.h> int a[105][105],f[105][105]; int max(int a,int b)