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 }
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时间： 2024-10-20 01:01:49