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  • POJ 1039 Pipe | 线段相交

    题目:

    给一个管子,有很多转弯处,问从管口的射线射进去最长能射到多远


    题解:

    根据黑书,可以证明的是这条光线一定经过了一个上顶点和下顶点

    所以我们枚举每对上下顶点就可以了

    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    #define eps 1e-5
    using namespace std;
    bool dcmp(double x,double y)
    {
        if (fabs(x-y)>eps) return 1;
        return 0;
    }
    struct point
    {
        double x,y;
        point () {};
        point (double _x,double _y)
    	{
    	    x=_x,y=_y;
    	}
        point operator + (const point &a)const
    	{
    	    return point(x+a.x,y+a.y);
    	}
        point operator - (const point &a) const
    	{
    	    return point(x-a.x,y-a.y);
    	}
        double operator * (const point &a)const
    	{
    	    return x*a.y-a.x*y;
    	}
        double dot (const point &a,const point &b)
    	{
    	    return a.x*b.x+a.y*b.y;
    	}
        bool operator < (const point &a)const
    	{
    	    return x<a.x;
    	}
        bool operator == (const point &a)const
    	{
    	    return dcmp(x,a.x) && dcmp(y,a.y);
    	}
    }up[105],down[105];
    inline double Max(double a,double b)
    {
        return a>b?a:b;
    }
    double Multi(point a,point b,point c)
    {
        return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
    }
    bool Across(point a,point b,point c,point d)
    {
        double tmp=Multi(c,b,a)*Multi(d,b,a);
        if (tmp<0 || fabs(tmp)<eps) return 1;
        return 0;
    }
    double getIntersect(point a,point b,point c,point d)
    {
        double A1=b.y-a.y,B1=a.x-b.x,C1=(b.x-a.x)*a.y-(b.y-a.y)*a.x;
        double A2=d.y-c.y,B2=c.x-d.x,C2=(d.x-c.x)*c.y-(d.y-c.y)*c.x;
        return (C2*B1-C1*B2)/(A1*B2-A2*B1);
    }
    int n,flag;
    double ans;
    int main()
    {
        while (scanf("%d",&n) && n)
        {
    	for (int i=0;i<n;i++)
    	{
    	    scanf("%lf%lf",&up[i].x,&up[i].y);
    	    down[i].x=up[i].x;
    	    down[i].y=up[i].y-1;
    	}
    	ans=up[0].x;
    	flag=0;
    	for (int i=0;i<n && !flag;i++)
    	    for (int j=0;j<n && !flag;j++)
    		if (i!=j)
    		{
    		    int k;
     		    for  (k=0;k<n;k++)
    			if (!Across(up[i],down[j],up[k],down[k])) break;
    		    if (k==n) flag=1;
    		    else if (k>i && k>j)
    		    {
    			double tmp;
    			tmp=getIntersect(up[i],down[j],up[k-1],up[k]);
    			//	printf("%.2f
    ",tmp);
    			if (tmp>ans) ans=tmp;
    			tmp=getIntersect(up[i],down[j],down[k-1],down[k]);
    			//	printf("%.2f
    ",tmp);
    			if (tmp>ans) ans=tmp;
    		    }
    		}
      	if (flag) puts("Through all the pipe.");
    	else printf("%.2f
    ",ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrsheep/p/8010574.html
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