The Doors(几何+最短路，好题)

The Doors

http://poj.org/problem?id=1556

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10466		Accepted: 3891

Description

You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path are always (0, 5) and (10, 5). There will also be from 0 to 18 vertical walls inside the chamber, each with two doorways. The figure below illustrates such a chamber and also shows the path of minimal length. 

Input

The input data for the illustrated chamber would appear as follows.

2 
4 2 7 8 9 
7 3 4.5 6 7

The first line contains the number of interior walls. Then there is a line for each such wall, containing five real numbers. The first number is the x coordinate of the wall (0 < x < 10), and the remaining four are the y coordinates of the ends of the doorways in that wall. The x coordinates of the walls are in increasing order, and within each line the y coordinates are in increasing order. The input file will contain at least one such set of data. The end of the data comes when the number of walls is -1.

Output

The output should contain one line of output for each chamber. The line should contain the minimal path length rounded to two decimal places past the decimal point, and always showing the two decimal places past the decimal point. The line should contain no blanks.

Sample Input

1
5 4 6 7 8
2
4 2 7 8 9
7 3 4.5 6 7
-1

Sample Output

10.00
10.06

Source

Mid-Central USA 1996

有18堵墙！！！因为没看清这个疯狂爆RE

poj上交C++会CE，要自己写hypot函数

double hypot(double x,double y){
    return sqrt(x*x+y*y);
}

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cmath>
 #include<vector>
 #include<algorithm>
 using namespace std;
 const double eps=1e-;
 const double INF=1e20;
 const double PI=acos(-1.0);
 const int maxp=;
 int sgn(double x){
     if(fabs(x)<eps) return ;
     if(x<) return -;
     else return ;
 }
 inline double sqr(double x){return x*x;}
 struct Point{
     double x,y;
     Point(){}
     Point(double _x,double _y){
         x=_x;
         y=_y;
     }
     void input(){
         scanf("%lf %lf",&x,&y);
     }
     void output(){
         printf("%.2f %.2f\n",x,y);
     }
     bool operator == (const Point &b)const{
         return sgn(x-b.x) ==  && sgn(y-b.y)== ;
     }
     bool operator < (const Point &b)const{
         return sgn(x-b.x)==?sgn(y-b.y)<:x<b.x;
     }
     Point operator - (const Point &b)const{
         return Point(x-b.x,y-b.y);
     }
     //叉积
     double operator ^ (const Point &b)const{
         return x*b.y-y*b.x;
     }
     //点积
     double operator * (const Point &b)const{
         return x*b.x+y*b.y;
     }
     //返回长度
     double len(){
         return hypot(x,y);
     }
     //返回长度的平方
     double len2(){
         return x*x+y*y;
     }
     //返回两点的距离
     double distance(Point p){
         return hypot(x-p.x,y-p.y);
     }
     Point operator + (const Point &b)const{
         return Point(x+b.x,y+b.y);
     }
     Point operator * (const double &k)const{
         return Point(x*k,y*k);
     }
     Point operator / (const double &k)const{
         return Point(x/k,y/k);
     }

     //计算pa和pb的夹角
     //就是求这个点看a,b所成的夹角
     ///LightOJ1202
     double rad(Point a,Point b){
         Point p=*this;
         return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
     }
     //化为长度为r的向量
     Point trunc(double r){
         double l=len();
         if(!sgn(l)) return *this;
         r/=l;
         return Point(x*r,y*r);
     }
     //逆时针转90度
     Point rotleft(){
         return Point(-y,x);
     }
     //顺时针转90度
     Point rotright(){
         return Point(y,-x);
     }
     //绕着p点逆时针旋转angle
     Point rotate(Point p,double angle){
         Point v=(*this) -p;
         double c=cos(angle),s=sin(angle);
         return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
     }
 };

 struct Line{
     Point s,e;
     Line(){}
     Line(Point _s,Point _e){
         s=_s;
         e=_e;
     }
     bool operator==(Line v){
         return (s==v.s)&&(e==v.e);
     }
     //根据一个点和倾斜角angle确定直线，0<=angle<pi
     Line(Point p,double angle){
         s=p;
         if(sgn(angle-PI/)==){
             e=(s+Point(,));
         }
         else{
             e=(s+Point(,tan(angle)));
         }
     }
     //ax+by+c=0;
     Line(double a,double b,double c){
         if(sgn(a)==){
             s=Point(,-c/b);
             e=Point(,-c/b);
         }
         else if(sgn(b)==){
             s=Point(-c/a,);
             e=Point(-c/a,);
         }
         else{
             s=Point(,-c/b);
             e=Point(,(-c-a)/b);
         }
     }
     void input(){
         s.input();
         e.input();
     }
     void adjust(){
         if(e<s) swap(s,e);
     }
     //求线段长度
     double length(){
         return s.distance(e);
     }
     //返回直线倾斜角 0<=angle<pi
     double angle(){
         double k=atan2(e.y-s.y,e.x-s.x);
         if(sgn(k)<) k+=PI;
         if(sgn(k-PI)==) k-=PI;
         return k;
     }
     //点和直线的关系
     //1 在左侧
     //2 在右侧
     //3 在直线上
     int relation(Point p){
         int c=sgn((p-s)^(e-s));
         if(c<) return ;
         else if(c>) return ;
         else return ;
     }
     //点在线段上的判断
     bool pointonseg(Point p){
         return sgn((p-s)^(e-s))==&&sgn((p-s)*(p-e))<=;
     }
     //两向量平行(对应直线平行或重合)
     bool parallel(Line v){
         return sgn((e-s)^(v.e-v.s))==;
     }
     //两线段相交判断
     //2 规范相交
     //1 非规范相交
     //0 不相交
     int segcrossseg(Line v){
         int d1=sgn((e-s)^(v.s-s));
         int d2=sgn((e-s)^(v.e-s));
         int d3=sgn((v.e-v.s)^(s-v.s));
         int d4=sgn((v.e-v.s)^(e-v.s));
         if((d1^d2)==-&&(d3^d4)==-) return ;
         return (d1==&&sgn((v.s-s)*(v.s-e))<=||
                 d2==&&sgn((v.e-s)*(v.e-e))<=||
                 d3==&&sgn((s-v.s)*(s-v.e))<=||
                 d4==&&sgn((e-v.s)*(e-v.e))<=);
     }
     //直线和线段相交判断
     //-*this line -v seg
     //2 规范相交
     //1 非规范相交
     //0 不相交
     int linecrossseg(Line v){
         int d1=sgn((e-s)^(v.s-s));
         int d2=sgn((e-s)^(v.e-s));
         if((d1^d2)==-) return ;
         return (d1==||d2==);
     }
     //两直线关系
     //0 平行
     //1 重合
     //2 相交
     int linecrossline(Line v){
         if((*this).parallel(v))
             return v.relation(s)==;
         return ;
     }
     //求两直线的交点
     //要保证两直线不平行或重合
     Point crosspoint(Line v){
         double a1=(v.e-v.s)^(s-v.s);
         double a2=(v.e-v.s)^(e-v.s);
         return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
     }
     //点到直线的距离
     double dispointtoline(Point p){
         return fabs((p-s)^(e-s))/length();
     }
     //点到线段的距离
     double dispointtoseg(Point p){
         if(sgn((p-s)*(e-s))<||sgn((p-e)*(s-e))<)
             return min(p.distance(s),p.distance(e));
         return dispointtoline(p);
     }
     //返回线段到线段的距离
     //前提是两线段不相交，相交距离就是0了
     double dissegtoseg(Line v){
         return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
     }
     //返回点P在直线上的投影
     Point lineprog(Point p){
         return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
     }
     //返回点P关于直线的对称点
     Point symmetrypoint(Point p){
         Point q=lineprog(p);
         return Point(*q.x-p.x,*q.y-p.y);
     }
 };

 Line L[];
 int n;

 bool Check(Line a,Line b){
     if(sgn((a.s-a.e)^(b.s-a.e))*sgn((a.s-a.e)^(b.e-a.e))>) return false;
     if(sgn((b.s-b.e)^(a.s-b.e))*sgn((b.s-b.e)^(a.e-b.e))>) return false;
     if(sgn(max(a.s.x,a.e.x)-min(b.s.x,b.e.x))>=&&sgn(max(b.s.x,b.e.x)-min(a.s.x,a.e.x))>=
      &&sgn(max(a.s.y,a.e.y)-min(b.s.y,b.e.y))>=&&sgn(max(b.s.y,b.e.y)-min(a.s.y,a.e.y))>=)
         return true;
     else return false;
 }

 double mp[][];

 int co;

 void panduan(Line a,int xx,int yy){
     if(a.s.y==||a.s.y==||a.e.y==||a.e.y==) return;
     for(int i=;i<co;i++){
         if(i!=(xx+)/&&i!=(yy+)/){
             if(Check(a,L[i])){
                 return;
             }
         }
     }
     //cout<<xx<<" "<<yy<<" "<<a.length()<<endl;

     mp[xx][yy]=mp[yy][xx]=a.length();
 }

 int main(){
     while(~scanf("%d",&n)){
         if(n==-) break;
         double x,y1,y2,y3,y4;
         co=;
         for(int i=;i<;i++){
             for(int j=;j<;j++){
                 mp[i][j]=INF;
             }
         }
         Point s,e;
         s.x=,s.y=;
         e.x=,e.y=;
         //起点为0,终点为co
         for(int i=;i<=n;i++){
             scanf("%lf %lf %lf %lf %lf",&x,&y1,&y2,&y3,&y4);
             L[co].s.x=x,L[co].s.y=,L[co].e.x=x,L[co++].e.y=y1;
             L[co].s.x=x,L[co].s.y=y2,L[co].e.x=x,L[co++].e.y=y3;
             L[co].s.x=x,L[co].s.y=y4,L[co].e.x=x,L[co++].e.y=;
         }

         Line tmp;
         int j;
         //不包括起点和终点的建图
         for(int i=;i<co;i++){
             for(int j=i+;j<co;j++){
                 tmp.s=L[i].s,tmp.e=L[j].s;
                 panduan(tmp,(i-)*+,(j-)*+);
                 tmp.s=L[i].s,tmp.e=L[j].e;
                 panduan(tmp,(i-)*+,(j-)*+);
                 tmp.s=L[i].e,tmp.e=L[j].s;
                 panduan(tmp,(i-)*+,(j-)*+);
                 tmp.s=L[i].e,tmp.e=L[j].e;
                 panduan(tmp,(i-)*+,(j-)*+);
             }
         }
         //加上起点和终点
         for(int i=;i<co;i++){
             tmp.s=s,tmp.e=L[i].s;
             panduan(tmp,,(i-)*+);
             tmp.s=s,tmp.e=L[i].e;
             panduan(tmp,,(i-)*+);
             tmp.s=e,tmp.e=L[i].s;
             panduan(tmp,(i-)*+,);
             tmp.s=e,tmp.e=L[i].e;
             panduan(tmp,(i-)*+,);
         }
         tmp.s=s,tmp.e=e;
         panduan(tmp,,);
         for(int k=;k<=;k++)
             for(int i=;i<=;i++)
                 for(int j=;j<=;j++)
                     if(mp[i][j]>mp[i][k]+mp[k][j]+eps)
                          mp[i][j]=mp[i][k]+mp[k][j];
         printf("%.2f\n",mp[][]);
     }
     return ;
 }