poj3449Geometric Shapes
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 }
