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;
}