POJ 1556 The Doors 线段交 dijkstra

Posted 2020-09-28 ( m Lweleth)

LINK

题意：在$10*10$的几何平面内，给出n条垂直x轴的线，且在线上开了两个口，起点为$(0, 5)$，终点为$(10, 5)$，问起点到终点不与其他线段相交的情况下的最小距离。

思路：将每个开口的两端点作为一个节点，再枚举点与点间能否直接到达（判相交），以此建图求最短路。

 

/** @Date    : 2017-07-11 16:17:31
  * @FileName: POJ 1556 线段交+dijkstra 计算几何.cpp
  * @Platform: Windows
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version : $Id$
  */
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <utility>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <stack>
#include <queue>
#include <math.h>
//#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const double INF = 0x7f;
const int N = 1e5+20;
const double eps = 1e-8;

struct Point
{
	double x, y;
	Point(){}
	Point(double xx, double yy){x = xx, y = yy;}
	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 cross(Point a, Point b)
{
	return a.x * b.y - a.y * b.x;
}

struct Line
{
	Point s, t;
	Line(){}
	Line(Point ss, Point tt){s = ss, t = tt;}
};

double distc(Point p1, Point p2)
{
	return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}
double xmult(Point p1, Point p2, Point p0)  
{  
    return cross(p1 - p0, p2 - p0);  
}  

//判两点在线段异侧,点在线段上返回0  
bool opposite_side(Point p1, Point p2, Line l)  
{  
    return xmult(l.s, p1, l.t)*xmult(l.s, p2, l.t) < -eps;  
}  

bool JudgeInter(Line a, Line b)//判断线段相交 不包括端点及重合
{
	return opposite_side(a.s, a.t, b) && opposite_side(b.s, b.t, a);
}

Line l[200];
Point p[200];
double mp[200][200];
double dis[200];
bool vis[200];

void dijkstra(int n)
{
	MMF(vis);
	MMI(dis);
	dis[0] = 0;
	vis[0] = 1;
	queue<int>q;
	q.push(0);
	while(!q.empty())
	{
		int nw = q.front();
		q.pop();
		for(int i = 0; i < n; i++)
		{
			double dic = dis[nw] + mp[nw][i];
			if(dis[i] > dic)
				dis[i] = dic;
		}
		double mi = INF;
		int np = 0;
		for(int i = 0; i < n; i++)
			if(!vis[i] && mi > dis[i])
				mi = dis[i], np = i;
		if(mi == INF)
			break;
		q.push(np);
		vis[np] = 1;
	}
}
int n;
int main()
{
	while(~scanf("%d", &n) && n!=-1)
	{
		int cntl = 0;
		int cntp = 0;
		p[cntp++] = Point(0, 5.0000);
		double x, a, b, c, d;
		for(int i = 0; i < n; i++)
		{
			scanf("%lf%lf%lf%lf%lf", &x, &a, &b, &c, &d);
			p[cntp++] = Point(x, 0.000);
			p[cntp++] = Point(x, a);
			l[cntl++] = Line(p[cntp - 2], p[cntp - 1]);

			p[cntp++] = Point(x, b);
			p[cntp++] = Point(x, c);
			l[cntl++] = Line(p[cntp - 2], p[cntp - 1]);

			p[cntp++] = Point(x, d);
			p[cntp++] = Point(x, 10.0000);
			l[cntl++] = Line(p[cntp - 2], p[cntp - 1]);
		}
		p[cntp++] = Point(10.0000, 5.0000);
		/////
		for(int i = 0; i < cntp; i++)//枚举任意两个点间是否能直接到达 
		{
			for(int j = 0; j < cntp; j++)
			{
				if(j == i)
					continue;
				Line tmp = Line(p[i], p[j]);
				int flag = 0;
				for(int k = 0; k < cntl; k++)
				{
					if(JudgeInter(tmp, l[k]))//判断是否与线段相交
					{
						mp[i][j] = INF;
						flag = 1;
						break;
					}
				}

				if(!flag)
					mp[i][j] = distc(p[i], p[j]);
			}
		}
		dijkstra(cntp);

		/*for(int i = 0; i < cntp; i++)
			cout << dis[i] << endl;*/
		printf("%.2lf\n", dis[cntp - 1]);
	}
    return 0;
}