The Doors POJ1556 线段相交 最短路

题目链接

解法：暴力判断任意两点是否可直达，并求出距离

处理完后用floyd算法求任意两点最短路

输出dis[0][4*n+1]即可

#include  <map>
#include  <set>
#include  <cmath>
#include  <queue>
#include  <cstdio>
#include  <vector>
#include  <climits>
#include  <cstring>
#include  <cstdlib>
#include  <iostream>
#include  <algorithm> 
#include <stdio.h>
using namespace std;

const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){
	if(fabs(x) < eps)return 0;
	if(x < 0)return -1;
	else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}

double hypot(double a,double b)
{
	return sqrt(sqr(a)+sqr(b));
}
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 == (Point b)const{
		return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
	}
	bool operator < (Point b)const{
		return sgn(x-b.x)== 0?sgn(y-b.y)<0: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);//库函数
	}
	//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`
	//`测试 LightOJ1203`
	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);
	}
	//`逆时针旋转90度`
	Point rotleft(){
		return Point(-y,x);
	}
	//`顺时针旋转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/2) == 0){
			e = (s + Point(0,1));
		}
		else{
			e = (s + Point(1,tan(angle)));
		}
	}
	//ax+by+c=0
	Line(double a,double b,double c){
		if(sgn(a) == 0){
			s = Point(0,-c/b);
			e = Point(1,-c/b);
		}
		else if(sgn(b) == 0){
			s = Point(-c/a,0);
			e = Point(-c/a,1);
		}
		else{
			s = Point(0,-c/b);
			e = Point(1,(-c-a)/b);
		}
	}
	void input(){
		s.input();
		e.input();
	}
	void output()
	{
		s.output();
		e.output();
	}
	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) < 0)k += pi;
		if(sgn(k-pi) == 0)k -= pi;
		return k;
	}
	//`点和直线关系`
	//`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 < 0)return 1;
		else if(c > 0)return 2;
		else return 3;
	}
	// 点在线段上的判断
	bool pointonseg(Point p){
		return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
	}
	//`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)) == 0;
	}
	//`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)==-2 && (d3^d4)==-2 )return 2;
		return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
			(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
			(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
			(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
	}
	//`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)==-2) return 2;
		return (d1==0||d2==0);
	}
	//`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)==3;
		return 2;
	}
	//`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))<0 || sgn((p-e)*(s-e))<0)
			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(2*q.x-p.x,2*q.y-p.y);
	}
};

int n;
double a[20][6];
Point P[200];
Line L[200];

double dis[200][200];
int ok(Point a,Point b,int j,int i) 
{
	int flag=0;
	Line t=Line(a,b);
	for(int ii=j;ii<i;ii++) 
	{					
		if(t.segcrossseg(L[ii*4+1])||t.segcrossseg(L[ii*4+2]))
		{
			flag++;
			continue;						
		}
		else return 0;
	}
	if(flag==i-j) return 1;
	else return 0;
}

double judge(Point a,Point b,int j,int i) 
{
	if(ok(a,b,j,i))
	{
		return a.distance(b);
	}
	else return (double)1e9;
}
int main()
{
	Point S=Point(0,5);
	Point E=Point(10,5);
	while(~scanf("%d",&n))
	{
//		printf("%d\n",n);
		if(n==-1) return 0;
		memset(dp,(double)1e9,sizeof(dp));
		memset(dis,(double)1e9,sizeof(dis));
		for(int j=0;j<n;j++)
		for(int i=0;i<5;i++)
		{
			scanf("%lf",&a[j][i]);
			if(i)
			{
				P[j*4+i]=Point(a[j][0],a[j][i]);
				if(i==2)
				{
					L[j*4+1]=Line(P[j*4+1],P[j*4+2]);
				}
				if(i==4)
				{
					L[j*4+2]=Line(P[j*4+3],P[j*4+4]);
				}
			}
		}
		P[0]=S;
		P[4*n+1]=E;
		for(int i=0;i<=1+4*n;i++)
		{
			int ij=(i-1)/4;
			if(i)dis[i][0]=dis[0][i]=judge(P[0],P[i],0,ij);
			if(i!=4*n+1)dis[1+4*n][i]=dis[i][1+4*n]=judge(P[i],P[4*n+1],ij+1,n);
		}
		
		for(int i=0;i<=4*n+1;i++) dis[i][i]=0;
		
		for(int i=1;i<=4*n;i++)
		{
			for(int j=i+1;j<=4*n;j++)
			{
				dis[j][i]=dis[i][j]=judge(P[i],P[j],(i-1)/4+1,(j-1)/4+1);
			}
		}
		for(int i=0;i<=4*n+1;i++)//floyd 
		{
			for(int j=0;j<=4*n+1;j++)
			{
				for(int l=0;l<=4*n+1;l++)
				{
					dis[j][l]=min(dis[j][l],dis[j][i]+dis[i][l]);
				}
			}
		}
		printf("%.2lf\n",dis[0][4*n+1]);

	}
}