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

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 #include<map>
 using namespace std;
 #define N 600
 #define LL long long
 #define INF 0xfffffff
 #define zero(x) (((x)>0?(x):-(x))<eps)
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 map<string,int>f;
 vector<int>ed[];
 int g;
 struct point
 {
     double x,y;
     point(double x=,double y=):x(x),y(y) {}
 } p[];
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return pointt(a.x-b.x,a.y-b.y);
 }
 struct line
 {
     pointt u,v;
     int flag;
     char c;
 } li[N];
 vector<line>dd[];
 char s1[],s2[],s[];
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 point rotate(point a,double rad)
 {
     return point(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));
 }
 double dot(point a,point b)
 {
     return a.x*b.x+a.y*b.y;
 }
 double dis(point a)
 {
     return sqrt(dot(a,a));
 }
 double angle(point a,point b)
 {
     return acos(dot(a,b)/dis(a)/dis(b));
 }
 double cross(point a,point b)
 {
     return a.x*b.y-a.y*b.x;
 }
 
 double xmult(point p1,point p2,point p0)
 {
     return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
 }
 //判三点共线
 int dots_inline(point p1,point p2,point p3)
 {
     return zero(xmult(p1,p2,p3));
 }
 
 //判点是否在线段上,包括端点
 int dot_online_in(point p,point l1,point l2)
 {
     return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;
 }
 
 //判两点在线段同侧,点在线段上返回0
 
 int same_side(point p1,point p2,point l1,point l2)
 {
     return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;
 }
 
 //判两线段相交,包括端点和部分重合
 
 int intersect_in(point u1,point u2,point v1,point v2)
 {
     if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2))
         return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);
     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);
 }
 void init(int kk,char c)
 {
     int i;
     int  k = c-'A';
     if(kk==)
     {
         for(i = ; i <=  ; i+=)
         {
             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
         }
         p[].x=(p[].x+p[].x+p[].y-p[].y)/;
         p[].y=(p[].y+p[].y+p[].x-p[].x)/;
         p[].x=(p[].x+p[].x-p[].y+p[].y)/;
         p[].y=(p[].y+p[].y-p[].x+p[].x)/;
         p[] = p[];
         for(i = ; i <  ; i++)
         {
             li[++g].u = p[i];
             li[g].v = p[i+];
             dd[k].push_back(li[g]);
         }
     }
     else if(kk==)
     {
         for(i = ; i <=  ; i++)
         {
             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
         }
         point pp = point((p[].x+p[].x),(p[].y+p[].y));
         p[] = point(pp.x-p[].x,pp.y-p[].y);
         //printf("%.3f %.3f\n",p[4].x,p[4].y);
         p[] = p[];
         for(i = ; i <= ; i++)
         {
             li[++g].u = p[i];
             li[g].v = p[i+];
             li[g].c = c;
             dd[k].push_back(li[g]);
         }
     }
     else if(kk==)
     {
         for(i = ; i <= ; i++)
             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
         li[++g].u = p[];
         li[g].v = p[];
         li[g].c = c;
         dd[k].push_back(li[g]);
     }
     else if(kk==)
     {
         for(i = ; i <=  ; i++)
             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
         p[] = p[];
         for(i = ; i <=  ; i++)
         {
             li[++g].u = p[i];
             li[g].v = p[i+];
             li[g].c = c;
             dd[k].push_back(li[g]);
         }
     }
     else if(kk==)
     {
         int n;
         scanf("%d",&n);
         for(i = ; i <= n ; i++)
             scanf(" (%lf,%lf)",&p[i].x,&p[i].y);
         p[n+] = p[];
         for(i = ; i <= n ; i++)
         {
             li[++g].u= p[i];
             li[g].v = p[i+];
             li[g].c = c;
             dd[k].push_back(li[g]);
         }
     }
 }
 
 int main()
 {
     f["square"] = ;
     f["rectangle"] = ;
     f["line"] = ;
     f["triangle"] = ;
     f["polygon"] = ;
     int i,j,k;
     while(scanf("%s",s1)!=EOF)
     {
         if(s1[]=='.') break;
         if(s1[]=='-') continue;
         for(i = ; i <  ; i++)
         {
             ed[i].clear();
             dd[i].clear();
         }
         g = ;
         k=;
         scanf("%s",s2);
         s[++k] = s1[];
         init(f[s2],s1[]);
         while(scanf("%s",s1)!=EOF)
         {
             if(s1[]=='-') break;
             //cout<<s1<<endl;
             scanf("%s",s2);
             s[++k] = s1[];
             init(f[s2],s1[]);
         }
         //cout<<g<<endl;
         sort(s+,s+k+);
         for(i =  ; i <= k; i++)
         {
             int u,v;
             u = s[i]-'A';
             //cout<<u<<" "<<dd[u].size()<<endl;
             for(j = i+; j <= k ; j++)
             {
                 v = s[j]-'A';
                 int flag = ;
                 for(int ii =  ; ii < dd[u].size() ; ii++)
                 {
                     for(int jj =  ; jj < dd[v].size() ; jj++)
                     {
                         if(intersect_in(dd[u][ii].u,dd[u][ii].v,dd[v][jj].u,dd[v][jj].v))
                         {
 
                             flag = ;
                             break;
                         }
 //                    if(u==5&&v==22)
 //                        {
 //                            output(dd[u][ii].u);
 //                        output(dd[u][ii].v);
 //                        output(dd[v][jj].u);
 //                        output(dd[v][jj].v);
 //                        }
                     }
                     if(flag) break;
                 }
                 if(flag)
                 {
                     ed[u].push_back(v);
                     ed[v].push_back(u);
                 }
             }
         }
         for(i =  ; i <= k; i++)
         {
             int u = s[i]-'A';
             if(ed[u].size()==)
                 printf("%c has no intersections\n",s[i]);
             else
             {
 
                 sort(ed[u].begin(),ed[u].end());
                 if(ed[u].size()==)
                     printf("%c intersects with %c\n",s[i],ed[u][]+'A');
                 else if(ed[u].size()==)
                     printf("%c intersects with %c and %c\n",s[i],ed[u][]+'A',ed[u][]+'A');
                 else
                 {
                     printf("%c intersects with ",s[i]);
                     for(j =  ; j < ed[u].size()- ; j++)
                         printf("%c, ",ed[u][j]+'A');
                     printf("and %c\n",ed[u][j]+'A');
                 }
             }
         }
         puts("");
     }
     return ;
 }