题目大意：
这题不好描述，直接看原题吧……
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<b.pos;
}
}np[N*6];
set<L> 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<na) swap(na,nb);
block[i] = L(na,nb);
np[++m] = (node){na,i} ;
np[++m] = (node){nb,-i} ;
np[++m] = (node){P (na[0]+leth,na[1]),0};
np[++m] = (node){P (nb[0]+leth,nb[1]),0};
np[++m] = (node){P (na[0]-leth,na[1]),0};
np[++m] = (node){P (nb[0]-leth,nb[1]),0};
}
sort(np+1,np+m+1);
up.clear();
down.clear();
double ans = 0.0,tmp,ret=0.0;
j=1;
while (j<=6*n) {
if (np[j].id<0) {
if (block[-np[j].id][0][1]<-eps)
down.erase(block[-np[j].id]);
else up.erase(block[-np[j].id]);
}else if (np[j].id>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);
}
}