POJ 1556 线段相交+最短路

题意：给出一个房间的俯视图 求从（0，5）到（10，5）的最短路径。

求出没有墙挡住的两点距离再用弗洛伊德求出0-sum的最短路径就可以。需要注意最短路可能经过某墙的端点，这时候不能判断条路径为非法。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef double PointType;
#define oo 999999
struct point
{
    PointType x,y;
};
PointType Direction(point pi,point pj,point pk) //判断向量PiPj在向量PiPk的顺逆时针方向 +顺-逆0共线
{
    return (pj.x-pi.x)*(pk.y-pi.y)-(pk.x-pi.x)*(pj.y-pi.y);
}
bool On_Segment(point pi,point pj,point pk)
{
    if(pk.x>=min(pi.x,pj.x)&&pk.x<=max(pi.x,pj.x)&&pk.y>=min(pi.y,pj.y)&&pk.y<=max(pi.y,pj.y))
        return 1;
    return 0;
}
bool Segment_Intersect(point p1,point p2,point p3,point p4)
{
    PointType d1=Direction(p3,p4,p1),d2=Direction(p3,p4,p2),d3=Direction(p1,p2,p3),d4=Direction(p1,p2,p4);
    if(((d1>0&&d2<0)||(d1<0&&d2>0))&&((d3>0&&d4<0)||(d3<0&&d4>0)))
        return 1;
    return 0;
}
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double dist[100][100];
bool judge(point a,point b,point c)
{
    if(a.x!=c.x&&b.x!=c.x)
        return 1;
    return 0;
}
int main()
{
    int n;
    point data[80],edge[18][6];
    while(~scanf("%d",&n),n!=-1)
    {
        for(int i=0; i<100; i++)
            for(int j=0; j<100; j++)
                dist[i][j]=oo;
        int sum=1;
        double w[4],x;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf%lf%lf",&x,&w[0],&w[1],&w[2],&w[3]);
            for(int j=0; j<4; j++)
            {
                data[sum+j].x=x,data[sum+j].y=w[j];
                edge[i][1+j].x=x,edge[i][1+j].y=w[j];
            }
            edge[i][0].x=x,edge[i][0].y=0;
            edge[i][5].x=x,edge[i][5].y=10;
            sum+=4;
        }
        data[0].x=0,data[0].y=5;
        data[sum].x=10,data[sum].y=5;
        for(int i=0; i<=sum; i++)
            for(int j=0; j<=sum; j++)
            {
                if(data[i].x==data[j].x)
                    continue;
                int flag=1;
                for(int k=0; k<n; k++)
                {
                    if(!judge(data[i],data[j],edge[k][0]))
                        continue;
                    bool a1=Segment_Intersect(data[i],data[j],edge[k][0],edge[k][1]),
                            a2=Segment_Intersect(data[i],data[j],edge[k][2],edge[k][3]),
                               a3=Segment_Intersect(data[i],data[j],edge[k][4],edge[k][5]);
                    if(a1||a2||a3)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag)
                    dist[i][j]=dis(data[i],data[j]);
            }
        for(int k=0; k<=sum; ++k)
            for(int i=0; i<=sum; ++i)
                for(int j=0; j<=sum; ++j)
                    if(dist[i][k]+dist[k][j]<dist[i][j])
                        dist[i][j]=dist[i][k]+dist[k][j];
        printf("%.2f\n",dist[0][sum]);
    }
    return 0;
}