POJ 1556

POJ 1556 - The Doors
题意：
    在 10x10 的空间里有很多垂直的墙，不能穿墙，问你从(0,5) 到 (10,5)的最短距离是多少.
    
分析：
    
    要么直达，要么一定是墙的边缘点之间以及起始点、终点的连线.
    
    所以先枚举墙上每一点到其他点的直线可达距离，就是要判定该线段是否与墙相交(不含端点).
    
    然后最短路.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
const double INF = 1e10;
const double eps = 1e-10;
int dcmp(double x)
{
    return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1);
}
struct Point
{
    double x,y;
    Point(double x1 = 0,double y1 = 0) : x(x1), y(y1) {}
};
Point operator - (Point a,Point b)
{
    return Point(a.x - b.x, a.y - b.y);
}
double Det(Point a,Point b)
{
    return a.x * b.y - a.y * b.x; 
}
double Dot(Point a,Point b)
{
    return a.x * b.x + a.y * b.y;
}
double Length(Point a)
{
    return sqrt(Dot(a, a));
}
bool OnSegment(Point p, Point a1, Point a2)
{
    return dcmp(Det(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p) ) <= 0;
}
struct Line
{
    Point s,e;
    Line() {}
    Line(Point s1, Point e1) : s(s1), e(e1) {}
};
bool SegCross(Point a1, Point a2, Point b1, Point b2)
{
    double c1 = Det(a2 - a1, b1 - a1);
    double c2 = Det(a2 - a1, b2 - a1);
    double c3 = Det(b2 - b1, a1 - b1);
    double c4 = Det(b2 - b1, a2 - b1);
    if(dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0) return 1;
    else return 0;
}

int n;
Line l[100];
Point p[100];
double map[100][100];
int main()
{
    while (~scanf("%d", &n) && n != -1)
    {
        p[0] = Point(0, 5);
        p[1] = Point(10, 5);
        int t = 2, m = 0;
        for (int i = 1; i <= n; i++)
        {
            double x; Point p1,p2,p3,p4;
            scanf("%lf%lf%lf%lf%lf",&x, &p1.y, &p2.y, &p3.y, &p4.y);
            p1.x = p2.x = p3.x = p4.x = x;
            l[m++] = Line(Point(x,0),p1);
            l[m++] = Line(p2, p3);
            l[m++] = Line(p4, Point(x,10));
            p[t++] = p1; p[t++] = p2; p[t++] = p3; p[t++] = p4;
        }
        for (int i = 0; i < t; i++)
            for (int j = 0; j < t; j++)
                map[i][j] = INF;
        for (int i = 0; i < t; i++)
        {
            for (int j = i+1; j < t; j++)
            {
                if (p[i].x == p[j].x) continue;
                bool flag = 1; 
                for (int k = 0; k < m; k++)//枚举墙
                {
                    if(SegCross(p[i], p[j], l[k].e, l[k].s)) 
                    {
                        flag = 0; break;
                    }
                } 
                if (flag)
                {
                    map[i][j] = map[j][i] = Length(p[i]-p[j]);
                }
                    
            }
        }
        for(int i = 0; i < t; i++) map[i][i] = 0;
        for (int k = 0; k < t; k++)
            for (int i = 0; i < t;i++)
                for (int j = 0; j < t; j++)
                    map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
        printf("%.2f
",map[0][1]);
    }
}