USACO3.41Closed Fences(几何）

这题水的真不易。。300多行 累死了 对着数据查错啊 

枚举每个边上的点到源点 是否中间隔着别的边  每条边划分500份就够了  注意一下与源点在一条直线上的边不算

几何 啊，，好繁琐 参考各种模版。。

  1 /*

  2     ID: shangca2

  3     LANG: C++

  4     TASK: fence4

  5  */

  6 #include <iostream>

  7 #include<cstdio>

  8 #include<cstring>

  9 #include<algorithm>

 10 #include<stdlib.h>

 11 #include<cmath>

 12 using namespace std;

 13 typedef struct  pointt

 14 {

 15     double x,y;

 16     pointt(double x=0,double y=0):x(x),y(y){}

 17 }vec;

 18 struct node

 19 {

 20     int x1,y1,x2,y2,o1,o2;

 21 }po[210];

 22 vec p,pp[210],q[4010];

 23 int e,n;

 24 vec operator + (vec a,vec b){return vec(a.x+b.x,a.y+b.y);}

 25 vec operator - (pointt a,pointt b){return vec(a.x-b.x,a.y-b.y);}

 26 vec operator * (vec a,double b){return vec(a.x*b,a.y*b);}

 27 vec operator / (vec a,double b){return vec(a.x/b,a.y/b);}

 28 bool operator < (const pointt &a,const pointt &b)

 29 {

 30     return a.x<b.x||(a.x==b.x&&a.y<b.y);

 31 }

 32 const double eps = 1e-15;

 33 int dcmp(double x)

 34 {

 35     if(fabs(x)<eps) return 0;

 36     else return x<0?-1:1;

 37 }

 38 bool operator == (const pointt &a,const pointt &b)

 39 {

 40     return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;

 41 }

 42 double dot(vec a,vec b)

 43 {

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

 45 }

 46 double cross(vec a,vec b)

 47 {

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

 49 }

 50 bool segprointer(pointt a1,pointt a2,pointt b1,pointt b2)

 51 {

 52     double c1 = cross(a2-a1,b1-a1),c2 = cross(a2-a1,b2-a1),

 53            c3 = cross(b2-b1,a1-b1),c4 = cross(b2-b1,a2-b1);

 54     return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;

 55 }

 56 bool cmp(node a,node b)

 57 {

 58     if(a.o2==b.o2)

 59     return a.o1<b.o1;

 60     return a.o2<b.o2;

 61 }

 62 int judge(int i,int j)

 63 {

 64     int g;

 65 

 66     for(g = 1; g <= n ; g++)

 67     {

 68         if(g==i)

 69         continue;

 70         if(segprointer(q[j],p,pp[g],pp[g+1]))

 71             break;

 72     }

 73     if(g==n+1)

 74     {

 75         e++;

 76         if(i==n)

 77         {

 78             po[e].x1 = pp[i+1].x;

 79             po[e].y1 = pp[i+1].y;

 80             po[e].x2 = pp[i].x;

 81             po[e].y2 = pp[i].y;

 82             po[e].o1 = 1;

 83             po[e].o2 = i;

 84         }

 85         else

 86         {

 87             po[e].o1 = i;

 88             po[e].o2 = i+1;

 89             po[e].x1 = pp[i].x;

 90             po[e].y1 = pp[i].y;

 91             po[e].x2 = pp[i+1].x;

 92             po[e].y2 = pp[i+1].y;

 93         }

 94         return 1;

 95     }

 96     return 0;

 97 }

 98 double xmult(vec p1,vec p2,vec p3)

 99 {

100     return (p1.x-p3.x)*(p2.y-p3.y)-(p2.x-p3.x)*(p1.y-p3.y);

101 }

102 int dots(vec p1,vec p2,vec p3)

103 {

104     return dcmp(xmult(p1,p2,p3));

105 }

106 int main()

107 {

108     freopen("fence4.in","r",stdin);

109     freopen("fence4.out","w",stdout);

110     int i,j,o,g;

111     cin>>n;

112     cin>>p.x>>p.y;

113     for(i = 1 ;i <= n ; i++)

114     cin>>pp[i].x>>pp[i].y;

115     pp[n+1] = pp[1];

116     for(i = 3; i <= n ; i++)

117     {

118         for(j = 1; j <= i-3 ; j++)

119         {

120             if(i==n&&j==1)

121             continue;

122             if(segprointer(pp[i],pp[i+1],pp[j],pp[j+1]))

123             {

124                 puts("NOFENCE");

125                 return 0;

126             }

127         }

128     }

129     vec no;

130     for(i = 1; i <= n ; i++)

131     {

132         if(!dots(p,pp[i],pp[i+1]))

133         continue;

134         o = 0;

135         int fo=0;

136         if(pp[i+1].x==pp[i].x)

137         {

138             double x1 = pp[i].x;

139             double y1;

140             if(pp[i].y<pp[i+1].y)

141             {

142                 double xx = (pp[i+1].y-pp[i].y)/300.0;

143                 y1 = pp[i].y+xx;

144                 while(y1<pp[i+1].y)

145                 {

146                     o++;

147                     q[o].x = x1;

148                     q[o].y = y1;

149                     if(judge(i,o))

150                     {

151                         fo = 1;

152                         break;

153                     }

154                     y1+=xx;

155                 }

156             }

157             else

158             {

159                 double xx = (pp[i].y-pp[i+1].y)/300.0;

160                 y1 = pp[i].y-xx;

161                 while(y1>pp[i+1].y)

162                 {

163                     o++;

164                     q[o].x = x1;

165                     q[o].y = y1;

166                     if(judge(i,o))

167                     {

168                         fo = 1;

169                         break;

170                     }

171                     y1-=xx;

172                 }

173             }

174         }

175         else if(pp[i+1].y==pp[i].y)

176         {

177 

178             double x1;

179             double y1=pp[i].y;

180             if(pp[i+1].x>pp[i].x)

181             {

182                 double xx = (pp[i+1].x-pp[i].x)/500.0;

183                 x1 = pp[i].x+xx;

184                 while(x1<pp[i+1].x)

185                 {

186                     o++;

187                     q[o].x = x1;

188                     q[o].y = y1;

189                     if(judge(i,o))

190                     {

191                         fo = 1;

192                         break;

193                     }

194                     x1+=xx;

195                 }

196             }

197             else

198             {

199                 double xx = (pp[i].x-pp[i+1].x)/500.0;

200                 x1 = pp[i].x-xx;

201                 while(x1>pp[i+1].x)

202                 {

203                     o++;

204                     q[o].x = x1;

205                     q[o].y = y1;

206                     if(judge(i,o))

207                     {

208                         fo = 1;

209                         if(i==7)

210                         cout<<x1<<" "<<y1<<endl;

211                         break;

212                     }

213                     x1-=xx;

214                 }

215             }

216         }

217         else

218         {

219             double d = (pp[i+1].y-pp[i].y)/(pp[i+1].x-pp[i].x);

220             double  k1,x1,y1;

221             double xx = fabs((pp[i+1].x-pp[i].x)/500.0);

222             k1 = 1.0*fabs(d);

223             if(pp[i+1].x>pp[i].x&&pp[i+1].y>pp[i].y)

224             {

225                 x1 = pp[i].x+0.1;

226                 y1 = pp[i].y+0.1*k1;

227                 while(x1<pp[i+1].x&&y1<pp[i+1].y)

228                 {

229                     o++;

230                     q[o].x = x1;

231                     q[o].y = y1;

232                     if(judge(i,o))

233                     {

234                         fo = 1;

235                         break;

236                     }

237                     x1+=xx;

238                     y1+=xx*k1;

239                 }

240             }

241             else if(pp[i+1].x<pp[i].x&&pp[i+1].y>pp[i].y)

242             {

243                 x1 = pp[i].x-0.1;

244                 y1 = pp[i].y+0.1*k1;

245                 while(x1>pp[i+1].x&&y1<pp[i+1].y)

246                 {

247                     o++;

248                     q[o].x = x1;

249                     q[o].y = y1;

250                     if(judge(i,o))

251                     {

252                         fo = 1;

253                         break;

254                     }

255                     x1-=xx;

256                     y1+=xx*k1;

257                 }

258             }

259             else if(pp[i+1].x<pp[i].x&&pp[i+1].y<pp[i].y)

260             {

261                 x1 = pp[i].x-0.1;

262                 y1 = pp[i].y-0.1*k1;

263                 while(x1>pp[i+1].x&&y1>pp[i+1].y)

264                 {

265                     o++;

266                     q[o].x = x1;

267                     q[o].y = y1;

268                     if(judge(i,o))

269                     {

270                         fo = 1;

271                         break;

272                     }

273                     x1-=xx;

274                     y1-=xx*k1;

275                 }

276             }

277             else

278             {

279                 x1 = pp[i].x+0.1;

280                 y1 = pp[i].y-0.1*k1;

281                 while(x1<pp[i+1].x&&y1>pp[i+1].y)

282                 {

283                     o++;

284                     q[o].x = x1;

285                     q[o].y = y1;

286                     if(judge(i,o))

287                     {

288                         fo = 1;

289                         break;

290                     }

291                     x1+=xx;

292                     y1-=xx*k1;

293                 }

294             }

295         }

296     }

297     sort(po+1,po+e+1,cmp);

298     cout<<e<<endl;

299     for(i = 1; i <= e ; i++)

300     {

301         cout<<po[i].x1<<" "<<po[i].y1<<" "<<po[i].x2<<" "<<po[i].y2<<endl;

302     }

303     return 0;

304 }

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