最短路+线段交 POJ 1556 好题

// 最短路+线段交 POJ 1556 好题
// 题意：从(0,5)到(10,5)的最短距离，中间有n堵墙，每堵上有两扇门可以通过
// 思路：先存图。直接n^2来暴力，不好写。分成三部分，起点 终点和之间的点：中间点之间：起点和终点的距离
// n最大为18所以直接n^3最短路


#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
const double inf = 1e9;
const int N = 5000;
const double eps = 1e-8;
void fre() {freopen("in.txt","r",stdin);}

int sgn(double x){
    if(fabs(x)<eps) return  0;
    if(x<0) return -1;
    return 1;
}

struct Point{
    double x,y;
    Point(){}
    Point(double _x,double _y){
        x=_x;y=_y;
    }
    Point operator -(const Point &b)const{
        return Point(x-b.x,y-b.y);
    }
    double operator *(const Point &b)const{
        return x*b.x+y*b.y;
    }
    double operator ^(const Point &b)const{
        return x*b.y-y*b.x;
    } 
};

struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s,e=_e;
    }
};
Line line[N];
double xmult(Point p0,Point p1,Point p2){
    return (p1-p0)^(p2-p0);
}

double dist(Point a,Point b){
    return sqrt((a-b)*(a-b));
}

bool inter(Line l1,Line l2){
    return max(l1.s.x,l1.e.x)>=min(l2.s.x,l2.e.x)&&max(l2.s.x,l2.e.x)>=min(l1.s.x,l1.e.x)&&max(l1.s.y,l1.e.y)>=min(l2.s.y,l2.e.y)&&max(l2.s.y,l2.e.y)>=min(l1.s.y,l1.e.y)&&sgn((l2.s-l1.e)^(l1.s-l1.e))*sgn((l2.e-l1.e)^(l1.s-l1.e))<=0&&sgn((l1.s-l2.e)^(l2.s-l2.e))*sgn((l1.e-l2.e)^(l2.s-l2.e))<=0;
}

double d[100][100];
int main(){
    // fre();
    int n;
    while(~scanf("%d",&n)){
        if(n==-1) break;
        // clc(d,
        for(int i=0;i<=4*n+1;i++){
            for(int j=0;j<=4*n+1;j++){
                if(i==j) d[i][j]=0;
                else d[i][j]=inf;
            }
        }
        for(int i=1;i<=n;i++){
            double x,y3,y1,y4,y2;
            scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
            line[i*2-1]=Line(Point(x,y1),Point(x,y2));
            line[i*2]=Line(Point(x,y3),Point(x,y4));
        }
        bool flag;
        for(int i=1;i<=4*n;i++){
            flag=true;
            int cn=(i+3)/4;
            Point tem;
            if(i&1) tem=line[(i+1)/2].s;
            else tem=line[(i+1)/2].e;
            for(int j=1;j<cn;j++){
                if(inter(line[j*2-1],Line(Point(0,5),tem))==false&&inter(line[j*2],Line(Point(0,5),tem))==false){
                    flag=false;
                    break;
                }
            }
            if(flag) d[0][i]=d[i][0]=dist(Point(0,5),tem);
            flag=true;
            for(int j=cn+1;j<=n;j++){
                if(inter(line[j*2-1],Line(Point(10,5),tem))==false&&inter(line[j*2],Line(Point(10,5),tem))==false){
                    flag=false;
                    break;
                }
            }
            if(flag) d[4*n+1][i]=d[i][4*n+1]=dist(Point(10,5),tem);
        }
        flag=true;
        for(int i=1;i<=4*n;i++){
            for(int j=i+1;j<=4*n;j++){
               flag=true;
               Point p1,p2;
               int cn1,cn2;
               cn1=(i+3)/4;
               cn2=(j+3)/4;
               if(i&1) p1=line[(i+1)/2].s;
               else p1=line[(i+1)/2].e;
               if(j&1) p2=line[(j+1)/2].s;
               else p2=line[(j+1)/2].e;
               for(int cn=cn1+1;cn<cn2;cn++){
                    if(inter(line[cn*2-1],Line(p1,p2))==false&&inter(line[cn*2],Line(p1,p2))==false){
                      flag=false;
                      break;
                    }
               }
               if(flag) d[i][j]=d[j][i]=dist(p1,p2);
            }
        }
        flag=true;
        for(int i=1;i<=n;i++){
            if(inter(line[i*2-1],Line(Point(0,5),Point(10,5)))==false&&inter(line[i*2],Line(Point(0,5),Point(10,5)))==false){
                flag=false;
                break;
            }
        }
        if(flag) d[0][4*n+1]=d[4*n+1][0]=dist(Point(0,5),Point(10,5));
        for(int i=0;i<=4*n+1;i++){
            for(int j=0;j<=4*n+1;j++){
                for(int k=0;k<=4*n+1;k++){
                    if(d[i][k]+d[k][j]<d[i][j]) d[i][j]=d[i][k]+d[k][j];
                }
            }
        }
        printf("%.2f\n",d[0][4*n+1]);
    }
    return 0;
}