poj3449Geometric Shapes

链接

繁琐。

处理出来所有的线段，再判断相交。

对于正方形的已知对角顶点求剩余两顶点 （列出4个方程求解）

p[1].x=(p[0].x+p[2].x+p[2].y-p[0].y)/2;

p[1].y=(p[0].y+p[2].y+p[0].x-p[2].x)/2;

p[3].x=(p[0].x+p[2].x-p[2].y+p[0].y)/2;

p[3].y=(p[0].y+p[2].y-p[0].x+p[2].x)/2;

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<vector>

  7 #include<cmath>

  8 #include<queue>

  9 #include<set>

 10 #include<map>

 11 using namespace std;

 12 #define N 600

 13 #define LL long long

 14 #define INF 0xfffffff

 15 #define zero(x) (((x)>0?(x):-(x))<eps)

 16 const double eps = 1e-8;

 17 const double pi = acos(-1.0);

 18 const double inf = ~0u>>2;

 19 map<string,int>f;

 20 vector<int>ed[30];

 21 int g;

 22 struct point

 23 {

 24     double x,y;

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

 26 } p[600];

 27 typedef point pointt;

 28 pointt operator -(point a,point b)

 29 {

 30     return pointt(a.x-b.x,a.y-b.y);

 31 }

 32 struct line

 33 {

 34     pointt u,v;

 35     int flag;

 36     char c;

 37 } li[N];

 38 vector<line>dd[30];

 39 char s1[10],s2[15],s[30];

 40 int dcmp(double x)

 41 {

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

 43     return x<0?-1:1;

 44 }

 45 point rotate(point a,double rad)

 46 {

 47     return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));

 48 }

 49 double dot(point a,point b)

 50 {

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

 52 }

 53 double dis(point a)

 54 {

 55     return sqrt(dot(a,a));

 56 }

 57 double angle(point a,point b)

 58 {

 59     return acos(dot(a,b)/dis(a)/dis(b));

 60 }

 61 double cross(point a,point b)

 62 {

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

 64 }

 65 

 66 double xmult(point p1,point p2,point p0)

 67 {

 68     return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);

 69 }

 70 //判三点共线

 71 int dots_inline(point p1,point p2,point p3)

 72 {

 73     return zero(xmult(p1,p2,p3));

 74 }

 75 

 76 //判点是否在线段上,包括端点

 77 int dot_online_in(point p,point l1,point l2)

 78 {

 79     return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;

 80 }

 81 

 82 //判两点在线段同侧,点在线段上返回0

 83 

 84 int same_side(point p1,point p2,point l1,point l2)

 85 {

 86     return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;

 87 }

 88 

 89 //判两线段相交,包括端点和部分重合

 90 

 91 int intersect_in(point u1,point u2,point v1,point v2)

 92 {

 93     if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2))

 94         return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);

 95     return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2);

 96 }

 97 void init(int kk,char c)

 98 {

 99     int i;

100     int  k = c-'A';

101     if(kk==1)

102     {

103         for(i = 0; i <= 2 ; i+=2)

104         {

105             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);

106         }

107         p[1].x=(p[0].x+p[2].x+p[2].y-p[0].y)/2;

108         p[1].y=(p[0].y+p[2].y+p[0].x-p[2].x)/2;

109         p[3].x=(p[0].x+p[2].x-p[2].y+p[0].y)/2;

110         p[3].y=(p[0].y+p[2].y-p[0].x+p[2].x)/2;

111         p[4] = p[0];

112         for(i = 0; i < 4 ; i++)

113         {

114             li[++g].u = p[i];

115             li[g].v = p[i+1];

116             dd[k].push_back(li[g]);

117         }

118     }

119     else if(kk==2)

120     {

121         for(i = 1; i <= 3 ; i++)

122         {

123             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);

124         }

125         point pp = point((p[1].x+p[3].x),(p[1].y+p[3].y));

126         p[4] = point(pp.x-p[2].x,pp.y-p[2].y);

127         //printf("%.3f %.3f\n",p[4].x,p[4].y);

128         p[5] = p[1];

129         for(i = 1; i <= 4; i++)

130         {

131             li[++g].u = p[i];

132             li[g].v = p[i+1];

133             li[g].c = c;

134             dd[k].push_back(li[g]);

135         }

136     }

137     else if(kk==3)

138     {

139         for(i = 1; i <= 2; i++)

140             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);

141         li[++g].u = p[1];

142         li[g].v = p[2];

143         li[g].c = c;

144         dd[k].push_back(li[g]);

145     }

146     else if(kk==4)

147     {

148         for(i = 1; i <= 3 ; i++)

149             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);

150         p[4] = p[1];

151         for(i = 1; i <= 3 ; i++)

152         {

153             li[++g].u = p[i];

154             li[g].v = p[i+1];

155             li[g].c = c;

156             dd[k].push_back(li[g]);

157         }

158     }

159     else if(kk==5)

160     {

161         int n;

162         scanf("%d",&n);

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

164             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);

165         p[n+1] = p[1];

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

167         {

168             li[++g].u= p[i];

169             li[g].v = p[i+1];

170             li[g].c = c;

171             dd[k].push_back(li[g]);

172         }

173     }

174 }

175 

176 int main()

177 {

178     f["square"] = 1;

179     f["rectangle"] = 2;

180     f["line"] = 3;

181     f["triangle"] = 4;

182     f["polygon"] = 5;

183     int i,j,k;

184     while(scanf("%s",s1)!=EOF)

185     {

186         if(s1[0]=='.') break;

187         if(s1[0]=='-') continue;

188         for(i =0 ; i < 26 ; i++)

189         {

190             ed[i].clear();

191             dd[i].clear();

192         }

193         g = 0;

194         k=0;

195         scanf("%s",s2);

196         s[++k] = s1[0];

197         init(f[s2],s1[0]);

198         while(scanf("%s",s1)!=EOF)

199         {

200             if(s1[0]=='-') break;

201             //cout<<s1<<endl;

202             scanf("%s",s2);

203             s[++k] = s1[0];

204             init(f[s2],s1[0]);

205         }

206         //cout<<g<<endl;

207         sort(s+1,s+k+1);

208         for(i = 1 ; i <= k; i++)

209         {

210             int u,v;

211             u = s[i]-'A';

212             //cout<<u<<" "<<dd[u].size()<<endl;

213             for(j = i+1; j <= k ; j++)

214             {

215                 v = s[j]-'A';

216                 int flag = 0;

217                 for(int ii = 0 ; ii < dd[u].size() ; ii++)

218                 {

219                     for(int jj = 0 ; jj < dd[v].size() ; jj++)

220                     {

221                         if(intersect_in(dd[u][ii].u,dd[u][ii].v,dd[v][jj].u,dd[v][jj].v))

222                         {

223 

224                             flag = 1;

225                             break;

226                         }

227 //                    if(u==5&&v==22)

228 //                        {

229 //                            output(dd[u][ii].u);

230 //                        output(dd[u][ii].v);

231 //                        output(dd[v][jj].u);

232 //                        output(dd[v][jj].v);

233 //                        }

234                     }

235                     if(flag) break;

236                 }

237                 if(flag)

238                 {

239                     ed[u].push_back(v);

240                     ed[v].push_back(u);

241                 }

242             }

243         }

244         for(i = 1 ; i <= k; i++)

245         {

246             int u = s[i]-'A';

247             if(ed[u].size()==0)

248                 printf("%c has no intersections\n",s[i]);

249             else

250             {

251 

252                 sort(ed[u].begin(),ed[u].end());

253                 if(ed[u].size()==1)

254                     printf("%c intersects with %c\n",s[i],ed[u][0]+'A');

255                 else if(ed[u].size()==2)

256                     printf("%c intersects with %c and %c\n",s[i],ed[u][0]+'A',ed[u][1]+'A');

257                 else

258                 {

259                     printf("%c intersects with ",s[i]);

260                     for(j = 0 ; j < ed[u].size()-1 ; j++)

261                         printf("%c, ",ed[u][j]+'A');

262                     printf("and %c\n",ed[u][j]+'A');

263                 }

264             }

265         }

266         puts("");

267     }

268     return 0;

269 }

View Code