poj1039Pipe（直线交点、叉积）

链接

之前刷poj计划时刷过，不过也没什么印象了。打铁还是趁热，还没热起来就放弃了，前面算是做了无用功，有如胡乱的看解题报告一样。

题目应该是比较经典的集合入门题，黑书上有一部分核心讲解。

题目中的最优光线必是要经过端点，这个黑书上提到了，应该也可以想到，然后就可以枚举一上一下的端点，判断它最长能走到哪里，首先可以判断出它是否能进的去第i个入口，

这个可以通过叉积进行判断上下点是不是在这条光线异侧，然后求直线的交点。

有一份好的模板很重要~

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 55
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y(y) {} //构造函数 方便代码编写
 }p[N];
 typedef Point pointt;
 pointt operator + (Point a,Point b)
 {
     return Point(a.x+b.x,a.y+b.y);
 }
 pointt operator - (Point a,Point b)
 {
     return Point(a.x-b.x,a.y-b.y);
 }
 pointt operator * (Point a,double b)
 {
     return Point(a.x*b,a.y*b);
 }
 pointt operator / (Point a,double b)
 {
     return Point(a.x/b,a.y/b);
 }
 bool operator < (const Point &a,const Point &b)
 {
     return a.x<b.x||(a.x==b.x&&a.y<b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     else return x<?-:;
 }
 bool operator == (const Point &a,const Point &b)
 {
     return dcmp(a.x-b.x)==&&dcmp(a.y-b.y)==;
 }
 //求点积以及利用点积求长度和夹角的函数
 double dot(Point a,Point b)
 {
     return a.x*b.x+a.y*b.y;
 }
 double cross(Point a,Point b,Point c)//差乘判左右 a->b与a->c向量的关系
 {
     return (a.x-c.x)*(a.y-b.y)-(a.y-c.y)*(a.x-b.x);
 }
 bool intersection1(Point p1, Point p2, Point p3, Point p4, Point& p)      // 直线相交
 {
     double a1, b1, c1, a2, b2, c2, d;
     a1 = p1.y - p2.y;
     b1 = p2.x - p1.x;
     c1 = p1.x*p2.y - p2.x*p1.y;
     a2 = p3.y - p4.y;
     b2 = p4.x - p3.x;
     c2 = p3.x*p4.y - p4.x*p3.y;
     d = a1*b2 - a2*b1;
     if (!dcmp(d))    return false;
     p.x = (-c1*b2 + c2*b1) / d;
     p.y = (-a1*c2 + a2*c1) / d;
     return true;
 }
 int main()
 {
     int n,i,j,g;
     while(scanf("%d",&n)&&n)
     {
         for(i  = ; i <= n ; i++)
         {
             scanf("%lf%lf",&p[i].x,&p[i].y);
             p[i+n].x = p[i].x;
             p[i+n].y = p[i].y-1.0;
         }
         double ans  = -INF;
         int flag = ;
         for(i =  ; i <= n; i++)
         {
             for(j = n+ ; j <= *n ; j++)
             {
                 if(j==n+i) continue;
                 Point sg ;
                 for(g = ; g <= n ; g++)
                 {
                     int d1 = dcmp(cross(p[i],p[j],p[g]));
                     int d2 = dcmp(cross(p[i],p[j],p[g+n]));
                     if(g==&&d1*d2<=) continue;
                     if(g==&&d1*d2>) break;
                     if(d1*d2>)
                     {
                         if(intersection1(p[i],p[j],p[g-],p[g],sg))
                         {
                             ans = max(ans,sg.x);
                         }
                         if(intersection1(p[i],p[j],p[n+g-],p[n+g],sg))
                         {
                             ans = max(ans,sg.x);
                         }
                         break;
                     }
                 }
                 if(g>n)
                 {
                     flag = ;
                     break;
                 }
             }
             if(flag) break;
         }
         if(flag) puts("Through all the pipe.");
         else printf("%.2f\n",ans);
     }
     return ;
 }