直线相交+区间覆盖——poj2074

感觉计算几何好多细节啊

这题的细节：障碍物在马路下面，在房子上面，都不用算

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 205

typedef double db;
const db eps=1e-8;
const db pi=acos(-1);
int sign(db k){if (k>eps) return 1; else if (k<-eps) return -1; return 0;}
int cmp(db k1,db k2){return sign(k1-k2);}
struct point{
    db x,y;
    point(){}
    point(db x,db y):x(x),y(y){}
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
};
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
int intersect(db l1,db r1,db l2,db r2){
    if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
}
int checkSS(point k1,point k2,point k3,point k4){
    return sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
point getLL(point k1,point k2,point k3,point k4){
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}

int n;
point k1,k2,k3,k4,k5,k6,k7;//k1,k2:房子 k3,k4:大街  k5,k6:障碍物 
struct Seg{
    db l,r;
}seg[N];
int comp(Seg a,Seg b){return a.l<b.l;}

int main(){
    while(cin>>k1.x && sign(k1.x)){
        cin>>k2.x>>k1.y;k2.y=k1.y;
        cin>>k3.x>>k4.x>>k3.y;k4.y=k3.y;
        cin>>n;
        for(int i=1;i<=n;i++){
            seg[i].l=seg[i].r=0;
            scanf("%lf%lf%lf",&k5.x,&k6.x,&k5.y);
            k6.y=k5.y;
            if(k5.y>=k1.y || k5.y<=k3.y)continue;
        
            k7=getLL(k1,k6,k3,k4);
            seg[i].r=k7.x;
            
            k7=getLL(k2,k5,k3,k4);
            seg[i].l=k7.x;
        }
        sort(seg+1,seg+1+n,comp);
        for(int i=1;i<=n;i++){
            if(seg[i].l<k3.x)seg[i].l=k3.x;
            if(seg[i].r<k3.x)seg[i].r=k3.x;
            if(seg[i].l>k4.x)seg[i].l=k4.x;
            if(seg[i].r>k4.x)seg[i].r=k4.x;
        }
        
        db Max=-1e18,nowR=0;
        for(int i=1;i<=n;i++){
            if(seg[i].l>nowR)
                Max=max(Max,seg[i].l-nowR);
            nowR=max(nowR,seg[i].r);
        }
        Max=max(Max,k4.x-seg[n].r);
        if(sign(Max)<=0)puts("No View");
        else printf("%.2f\n",Max);
    } 
}