POJ 1039 Pipe

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这题做了整整一下午，被折磨得死去活来。

思路是枚举两个端点，然后如果跟管道拐口的线都有交点则可以穿过整个管道，我一开始枚举端点，然后跟拐口的线比，然后如果没有交点，那么就计算x的值，但是这样会有一个问题，那就是如果枚举的两个端点前面有线没有相交，那么他会计算x的值，导致x的值变小了，所以在判断点前的线时，单独判断就好了。

////  main.cpp//  Richard////  Created by 邵金杰 on 16/8/8.//  Copyright © 2016年 邵金杰. All rights reserved.//#include<cstdio>#include<iostream>#include<vector>using namespace std;#define INF 1e20#define EPS 1e-6struct Point{    double x,y;    Point(double x,double y): x(x),y(y) {}};struct Vector{    double x,y;    Vector(double x,double y): x(x),y(y) {}};struct Line{    Point a,b;    Line(Point a,Point b): a(a),b(b) {}};Vector operator - (Point p,Point q){    return Vector(p.x-q.x,p.y-q.y);}Vector operator * (double k,Vector p){    return Vector(k*p.x,k*p.y);}Point operator + (Point p,Vector q){    return Point(p.x+q.x,p.y+q.y);}double getX(double y,Point a,Point b){    return a.x+(b.x-a.x)*((y-a.y)/(b.y-a.y));}vector<Point> ve;vector<Line>  le;double operator ^ (Vector p,Vector q){    return p.x*q.y-p.y*q.x;}double cross(Vector p,Vector q){    return p.x*q.y-p.y*q.x;}bool IsZero(double x){    return -EPS<x&&x<EPS;}double intersect(Line l,Line m){    double x=cross(l.a-m.a,m.b-m.a)/2;    double y=cross(m.b-m.a,l.b-m.a)/2;    Point p=l.a+(x/(x+y))*(l.b-l.a);    return p.x;}void work(){    double Max=-INF;    int lenv=(int)ve.size();    int lenl=(int)le.size();    for(int i=0;i<lenv;i++)    {        int s;        if(i%2==0) s=i+2;        else s=i+1;        for(int j=s;j<lenv;j++)        {            bool success=true;            int p=i/2;            for(int k=0;k<p;k++)            {                if(cross(le[k].a-ve[i],ve[j]-ve[i])*cross(le[k].b-ve[i],ve[j]-ve[i])>EPS)                {                    success=false;                    break;                }            }            if(!success) continue;            bool ok=true;            for(int k=p;k<lenl;k++)            {                if(cross(le[k].a-ve[i],ve[j]-ve[i])*cross(le[k].b-ve[i],ve[j]-ve[i])>EPS)                {                    ok=false;                    if(cross(le[k].a-ve[i],ve[j]-ve[i])*cross(le[k-1].a-ve[i],ve[j]-ve[i])<=-EPS){                        Max=max(Max,intersect(Line(le[k].a,le[k-1].a),Line(ve[j],ve[i])));                    }                    else if(cross(le[k].b-ve[i],ve[j]-ve[i])*cross(le[k-1].b-ve[i],ve[j]-ve[i])<=-EPS){                        Max=max(Max,intersect(Line(le[k].b,le[k-1].b),Line(ve[j],ve[i])));                    }                    break;                }            }            if(ok) {printf("Through all the pipe.\n"); return ;}        }    }    printf("%.2f\n",Max);}int main(){    int n;    double x,y;    while(scanf("%d",&n)&&n)    {        ve.clear();        le.clear();        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&x,&y);            ve.push_back(Point(x,y));            ve.push_back(Point(x,y-1));            le.push_back(Line(Point(x,y),Point(x,y-1)));        }        if(n<3) printf("Through all the pipe.\n");        else work();    }    return 0;}

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