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\n",tmp); if (tmp>ans) ans=tmp; tmp=getIntersect(up[i],down[j],down[k-1],down[k]); // printf("%.2f\n",tmp); if (tmp>ans) ans=tmp; } } if (flag) puts("Through all the pipe."); else printf("%.2f\n",ans); } return 0; }