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-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):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 0;
    else return x<0?-1:1;
}
bool operator == (const Point &a,const Point &b)
{
    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
//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  = 1; 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 = 0;
        for(i = 1 ; i <= n; i++)
        {
            for(j = n+1 ; j <= 2*n ; j++)
            {
                if(j==n+i) continue;
                Point sg ;
                for(g = 1; 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==1&&d1*d2<=0) continue;
                    if(g==1&&d1*d2>0) break;
                    if(d1*d2>0)
                    {
                        if(intersection1(p[i],p[j],p[g-1],p[g],sg))
                        {
                            ans = max(ans,sg.x);
                        }
                        if(intersection1(p[i],p[j],p[n+g-1],p[n+g],sg))
                        {
                            ans = max(ans,sg.x);
                        }
                        break;
                    }
                }
                if(g>n)
                {
                    flag = 1;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) puts("Through all the pipe.");
        else printf("%.2f
",ans);
    }
    return 0;
}