题目大意：

　　这题不好描述，直接看吧……

题解：

　　很无脑的题……就是卡精度+难写。代码能力还是太差了。

　　其实可以直接用long double肝过去。但我的代码似乎太丑了，以至于跑得奇慢无比。

代码：

　　

#include "bits/stdc++.h"using namespace std;inline int read() {    int s=0,k=1;char ch=getchar();    while (ch<'0'|ch>'9') ch=='-'?k=-1:0,ch=getchar();    while (ch>47&ch<='9') s=s*10+(ch^48),ch=getchar();    return s*k;}#define double long double const double eps=1e-10;const int N=1e4+10;inline int dcmp(double a) {    if (a>eps) return 1;    if (a<-eps) return -1;    return 0; }struct P{    P(double _x=0.0,double _y=0.0):x(_x),y(_y){}    double x,y;    double &operator [] (int k) {return k?y:x;}    friend bool operator <(P a,P b) {        return dcmp(a[0]-b[0])==-1||(dcmp(a[0]-b[0])==0&&dcmp(a[1]-b[1])==-1);    }    friend double operator *(P a,P b) {        return a[0]*b[1]-b[0]*a[1];    }    friend P operator /(P a,double b) {        return P(a[0]/b,a[1]/b);    }    friend P operator + (P a,P b) {        return P(a[0]+b[0],a[1]+b[1]);    }    friend P operator * (P a,double b) {        return P(a[0]*b,a[1]*b);    }    friend double operator /(P a,P b) {        return a[0]*b[0]+b[1]*a[1];    }    inline double leth(){        return sqrt(x*x+y*y);    }};P operator -(P a,P b) {    return P(a[0]-b[0],a[1]-b[1]);}    struct L{    L(){}L(P _a,P _b):a(_a),b(_b),c(_b-_a),slope(atan2((a-b)[1],(a-b)[0])){}    P a,b,c;double slope;    P &operator[] (int k) {        if(!k) return a;        if (k==1) return b;        return c;    }    friend double touying (L x,L y) {        return x[2]/y[2]/(x[2].leth());    }    friend double touying (L x,P y) {        return x[2]/(y-x[0])/(x[2].leth());    }    friend bool operator <(L x,L y) {        double pos = max (x[0][0],y[0][0]);        return dcmp(fabs(x[0][1] + x[2][1] * (pos-x[0][0]) / ( x[1][0] - x[0][0] )) -                   fabs(y[0][1] + y[2][1] * (pos-y[0][0]) / ( y[1][0] - y[0][0] )) ) == -1;                      }}block[N],gui;inline bool Judge(L a,L b) {    return dcmp(b[2]*(a[0]-b[0]))==1;}inline double dis(P p,L l) {    return l[2]*(p-l[0])/l[2].leth();}double lft[N],lth[N],rght[N],rth[N];int n,leth,m;int pl[N],pr[N];struct node {    P pos;int id;    friend bool operator <(node a,node b) {        return a.pos
     
       up,down;double res[N*6];inline double calc (double x2,double x1,L l) {    P delta=l[2]*(x2-x1)/l[2][0];    return delta.leth();}int main(){    int T=read();    register int i,j;    while (T-- ) {        n=read();        for (i=1;i<=n;++i) {            double x1=read(),y1=read(),x2=read(),y2=read();            if (x1>x2) swap(x1,x2),swap(y1,y2);            if (x1==x2&&y1>y2) swap(y1,y2);            block[i]=L(P(x1,y1),P(x2,y2));        }        double x1=read(),y1=read(),x2=read(),y2=read();leth=read();        if (x1>x2) swap(x1,x2),swap(y1,y2);        if (x1==x2&&y1>y2) swap(y1,y2);        gui=L(P(x1,y1),P(x2,y2));        m=0;        for (i=1;i<=n;++i) {            P a=block[i][0],b=block[i][1];            P na,nb;            na[0] = touying(gui,a),na[1] = dis(a,gui);            nb[0] = touying(gui,b),nb[1] = dis(b,gui);            if (nb
      
       0){                if (block[np[j].id][0][1]<-eps)                     down.insert(block[np[j].id]);                else up.insert(block[np[j].id]);            }            ++j;            res[j]=0.0;            if (up.size()){                tmp = calc (np[j].pos[0],np[j-1].pos[0],*up.begin());                res[j] +=tmp;            }            if (down.size()) {                tmp = calc (np[j].pos[0],np[j-1].pos[0],*down.begin());                res[j] +=tmp;            }        }        for (i=j=1;i<=6*n&&j<=6*n;++i,ans-=res[i]) {            while (j<=6*n&&dcmp(np[j].pos[0]-np[i].pos[0]-leth)<=0) ans+=res[j++];            ret=max(ret,ans);        }        printf("%.15Lf\n",ret);    }}