POJ 1556 线段相交

给定一个内含阻碍墙的房间，你要编程找到一条从起点到终点的最短路。

判断线段相交的题目

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

const int n_max=105;
const int inf=1<<30;
const double precision=1e-6;

struct point
{
	double x,y;
	void add(double tx,double ty)
	{
		x=tx;
		y=ty;
	}
}p[n_max];

struct wall
{
	point low,hig;
	void add(double x1,double y1,double x2,double y2)
	{
		low.add(x1,y1);
		hig.add(x2,y2);
	}
}w[n_max];
int np,nw;
int poi[n_max][n_max];
int wal[n_max][n_max];
double d[n_max];

void init()
{
	int i;
	for(i=0;i<n_max;i++)
	{
		poi[i][0]=0;
		wal[i][0]=0;
	}
	np=0;
	nw=0;
	p[np].add((double)0,(double)5);
	poi[0][++poi[0][0]]=np;
	d[np++]=0;
}

void addpoint(int i,double x,double y)
{
	p[np].add(x,y);
	poi[i][ ++poi[i][0] ] = np;
	d[np++]=inf;
}

void addwall(int i,double x1,double y1,double x2,double y2)
{
	if(x1==x2&&y1==y2)
	{
		return ;
	}
	w[nw].add(x1,y1,x2,y2);
	wal[i][ ++wal[i][0] ] = nw++;
}

double dist(point a,point b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int dblcmp(double d)
{
	if(fabs(d)<precision)
	{
		return 0;
	}
	return (d>0)?1:-1;
}

double det(double x1,double y1,double x2,double y2)
{
	return x1*y2 - x2*y1;
}

double cross(point a,point b,point c)
{
	return det(b.x-a.x,b.y-a.y,c.x-a.x,c.y-a.y);
}

bool segcross(point a,point b,point c,point d)  //判断线段是否相交
{
	return ( dblcmp(cross(a,c,d)) ^ dblcmp(cross(b,c,d)) ) == -2 &&
			(dblcmp(cross(c,a,b)) ^ dblcmp(cross(d,a,b)) ) == -2;
}

bool judge(int i,int pi,int j,int pj)
{
	int k,u,v,pk,c;
	v=poi[i][pi];
	u=poi[j][pj];
	for(k=j+1;k<i;k++)
	{
		for(pk=1;pk<=wal[k][0];pk++)
		{
			c=wal[k][pk];
    		if(segcross(p[u],p[v],w[c].low,w[c].hig))
    		{
    			return false;
    		}
		}
	}
	return true;
}

int main()
{
	int n;
	while(~scanf("%d",&n),n!=-1)
	{
		init();
		int i,j,pi,pj;
		double x,y1,y2,y3,y4;
		for(i=1;i<=n;i++)
		{
			scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
			addpoint(i,x,y1);
			addpoint(i,x,y2);
			addpoint(i,x,y3);
			addpoint(i,x,y4);
			addwall(i,x,0,x,y1);
			addwall(i,x,y2,x,y3);
			addwall(i,x,y4,x,10);
		}
		addpoint(n+1,10,5);

		int u,v;
		for(i=1;i<=n+1;i++)
		{
			for(j=0;j<i;j++)
			{
				for(pi=1;pi<=poi[i][0];pi++)
				{
					for(pj=1;pj<=poi[j][0];pj++)
					{
						if(judge(i,pi,j,pj))
						{
							v=poi[i][pi];
							u=poi[j][pj];
							d[v] = min(d[v], d[u]+dist(p[u],p[v]) );
						}
					}
				}
			}
		}
		u=poi[n+1][1];
		printf("%.2lf\n",d[u]);
	}
	return 0;
}