POJ 2074 Line of Sight 计算几何

题意：

给出房子，障碍物，观光线（都为平行于x轴的线段）。问在观光线上能看到整个房子的最长距离

分析：

将房屋的端点与障碍物的端点连线，求出与观光线的横坐标。这些坐标会把观光线分成多个区间，然后枚举每一个区间的中点，来判断这个区间是否能看到整个房子

要注意的是：不一定每个障碍物都在房屋与观光线之间，与房屋或观光线在同一条直线的可以忽略掉

如果你用的G++，double输出格式为%f

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define eps 1e-8
#define INF 1e9
#define OK sgn(tmp-lx1)>0 && sgn(tmp-lx2<0)
using namespace std;

typedef struct Point
{
    double x,y;
    Point() {};
    Point(double xx,double yy)
    {
        x=xx;
        y=yy;
    }
} Vector;

struct Line
{
    Point p,q;
    Line() {};
    Line(Point pp,Point qq)
    {
        p=pp;
        q=qq;
    }
};

int sgn(double x)
{
    if(fabs(x)<eps) return 0;
    return x<0? -1:1;
}

double crs_prdct(Vector a,Vector b)
{
    return a.x*b.y-b.x*a.y;
}

Vector operator - (Point a,Point b)
{
    return Vector(a.x-b.x,a.y-b.y);
}

double calcu(Point a,Point b,double y)
{
    return (a.x-b.x)/(a.y-b.y)*(y-a.y)+a.x;
}

const int maxn=105;
Line hou,pro,obs[maxn];
double dot[maxn];

int main()
{
//    freopen("in.txt","r",stdin);
    int n,m;
    double x1,x2,y,lx1,lx2,ly;
    while(scanf("%lf%lf%lf",&x1,&x2,&y))
    {
        if(x1==0 && x2==0 && y==0) break;
        hou=Line(Point(x1,y),Point(x2,y));
        scanf("%lf%lf%lf",&lx1,&lx2,&ly);
        scanf("%d",&n);
        m=0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf%lf",&x1,&x2,&y);
            if(sgn(y-hou.p.y)>=0 || sgn(y-ly)<=0) continue;
            obs[m++]=Line(Point(x1,y),Point(x2,y));
        }
        int cnt=0;
        dot[cnt++]=lx1;
        double tmp;
        for(int i=0; i<m; i++)
        {
            if(obs[i].p.y>=hou.p.y || obs[i].p.y<ly) continue;
            tmp=calcu(hou.p,obs[i].p,ly);
            if(OK) dot[cnt++]=tmp;
            tmp=calcu(hou.p,obs[i].q,ly);
            if(OK) dot[cnt++]=tmp;
            tmp=calcu(hou.q,obs[i].p,ly);
            if(OK) dot[cnt++]=tmp;
            tmp=calcu(hou.q,obs[i].q,ly);
            if(OK) dot[cnt++]=tmp;
        }
        dot[cnt++]=lx2;
        sort(dot,dot+cnt);
        double ans=0;
        tmp=0;
        for(int i=0; i<cnt-1; i++)
        {
            bool flag=true;
            Point mid=Point((dot[i]+dot[i+1])/2,ly);
            for(int j=0; j<m; j++)
            {
                if(sgn(crs_prdct(hou.p-mid,obs[j].q-mid))<=0 && sgn(crs_prdct(hou.q-mid,obs[j].p-mid))>=0)
                {
                    flag=false;
                    break;
                }
            }
            if(flag)
            {
                tmp+=dot[i+1]-dot[i];
                ans=max(ans,tmp);
            }
            else tmp=0;
        }
        if(sgn(ans)>0) printf("%.2f
",ans);
        else puts("No View");
    }
    return 0;
}