我人傻了，我两份半平面交的板子都错了？？
然后粘了网友的板子过了？？？？
我他妈？？？？
这现场赛万一出个半平面交我都不知道粘哪份板子了
其实题目本身还挺简单的，
维护一下小凸包每条边切的点，
然后二分半平面交就行了。
代码几乎照着网友的抄的。。。。。。好像一共也就几行代码
#include <bits/stdc++.h>
#define mp make_pair
#define fi first
#define se second
#define pb push_back
using namespace std;
typedef double db;
const int maxn = 4e5+5;
const db eps=1e-13;
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);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内
struct point{
db x,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};}
int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
// 逆时针旋转
point turn(db k1){return (point){x*cos(k1)-y*sin(k1),x*sin(k1)+y*cos(k1)};}
point turn90(){return (point){-y,x};}
bool operator < (const point k1) const{
int a=cmp(x,k1.x);
if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
}
db abs(){return sqrt(x*x+y*y);}
db abs2(){return x*x+y*y;}
db dis(point k1){return ((*this)-k1).abs();}
point unit(){db w=abs(); return (point){x/w,y/w};}
void scan(){double k1,k2; scanf("%lf%lf",&k1,&k2); x=k1; y=k2;}
void print(){printf("%.11lf %.11lf\n",x,y);}
db getw(){return atan2(y,x);}
point getdel(){if (sign(x)==-1||(sign(x)==0&&sign(y)==-1)) return (*this)*(-1); else return (*this);}
int getP() const{return sign(y)==1||(sign(y)==0&&sign(x)==-1);}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));}
// -pi -> pi
int compareangle (point k1,point k2){//极角排序+
return k1.getP()<k2.getP()||(k1.getP()==k2.getP()&&sign(cross(k1,k2))>0);
}
point proj(point k1,point k2,point q){ // q 到直线 k1,k2 的投影
point k=k2-k1;return k1+k*(dot(q-k1,k)/k.abs2());
}
point reflect(point k1,point k2,point q){return proj(k1,k2,q)*2-q;}
int clockwise(point k1,point k2,point k3){// k1 k2 k3 逆时针 1 顺时针 -1 否则 0
return sign(cross(k2-k1,k3-k1));
}
int checkLL(point k1,point k2,point k3,point k4){// 求直线 (L) 线段 (S)k1,k2 和 k3,k4 的交点
return cmp(cross(k3-k1,k4-k1),cross(k3-k2,k4-k2))!=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 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 intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}
db disSP(point k1,point k2,point q){
point k3=proj(k1,k2,q);
if (inmid(k1,k2,k3)) return q.dis(k3); else return min(q.dis(k1),q.dis(k2));
}
db disSS(point k1,point k2,point k3,point k4){
if (checkSS(k1,k2,k3,k4)) return 0;
else return min(min(disSP(k1,k2,k3),disSP(k1,k2,k4)),min(disSP(k3,k4,k1),disSP(k3,k4,k2)));
}
int onS(point k1,point k2,point q){return inmid(k1,k2,q)&&sign(cross(k1-q,k2-k1))==0;}
struct circle{
point o; db r;
void scan(){o.scan(); scanf("%lf",&r);}
int inside(point k){return cmp(r,o.dis(k));}
};
struct line{
// p[0]->p[1]
point p[2];
line(point k1,point k2){p[0]=k1; p[1]=k2;}
line(){}
point& operator [] (int k){return p[k];}
int include(point k){return sign(cross(p[1]-p[0],k-p[0]))>0;}
point dir(){return p[1]-p[0];}
line push(){ // 向外 ( 左手边 ) 平移 eps
const db eps = 1e-6;
point delta=(p[1]-p[0]).turn90().unit()*eps;
return {p[0]-delta,p[1]-delta};
}
};
point getLL(line k1,line k2){return getLL(k1[0],k1[1],k2[0],k2[1]);}
int parallel(line k1,line k2){return sign(cross(k1.dir(),k2.dir()))==0;}
int sameDir(line k1,line k2){return parallel(k1,k2)&&sign(dot(k1.dir(),k2.dir()))==1;}
int operator < (line k1,line k2){
if (sameDir(k1,k2)) return k2.include(k1[0]);
return compareangle(k1.dir(),k2.dir());
}
int checkpos(line k1,line k2,line k3){return k3.include(getLL(k1,k2));}
vector<line> getHL(vector<line> &L){ // 求半平面交 , 半平面是逆时针方向 , 输出按照逆时针
sort(L.begin(),L.end()); deque<line> q;
for (int i=0;i<(int)L.size();i++){
if (i&&sameDir(L[i],L[i-1])) continue;
while (q.size()>1&&!checkpos(q[q.size()-2],q[q.size()-1],L[i])) q.pop_back();
while (q.size()>1&&!checkpos(q[1],q[0],L[i])) q.pop_front();
q.push_back(L[i]);
}
while (q.size()>2&&!checkpos(q[q.size()-2],q[q.size()-1],q[0])) q.pop_back();
while (q.size()>2&&!checkpos(q[1],q[0],q[q.size()-1])) q.pop_front();
vector<line>ans; for (int i=0;i<q.size();i++) ans.push_back(q[i]);
return ans;
}
vector<point> ConvexHull(vector<point>A,int flag=1){ // flag=0 不严格 flag=1 严格
int n=A.size(); vector<point>ans(n*2);
sort(A.begin(),A.end()); int now=-1;
for (int i=0;i<A.size();i++){
while (now>0&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
ans[++now]=A[i];
} int pre=now;
for (int i=n-2;i>=0;i--){
while (now>pre&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
ans[++now]=A[i];
} ans.resize(now); return ans;
}
bool cclock(vector<point> &p){
int n = p.size();
for(int i=1;i<n-1;i++){
if(sign(cross(p[i]-p[i-1],p[i+1]-p[i-1]))>0) return 0;
else if(sign(cross(p[i]-p[i-1],p[i+1]-p[i-1]))<0){
reverse(p.begin(),p.end());
return 1;
}
}
return 1;
}
//117-151 更可靠的半平面交板子 from matthew99 upd 9-29
bool on_left(line l,point p){
return sign(cross(l.dir(),p-l[0]))>0;
}
inline bool half_plane(line *plane, const int &n_plane, point *poly, int &n_poly){
static int pos[maxn + 5];
static double ang[maxn + 5];
for(int i=0;i<n_plane;i++) ang[i] = atan2(plane[i].dir().y, plane[i].dir().x), pos[i] = i;
sort(pos, pos + n_plane, [&](int x, int y) { return ang[x] < ang[y]; });
int head, rear;
static point p[maxn + 5];
static line q[maxn + 5];
head = rear = 0;
q[rear++] = plane[pos[0]];
for(int i=0;i<n_plane;i++){
while (head < rear - 1 && !on_left(plane[pos[i]], p[rear - 2])) --rear;
while (head < rear - 1 && !on_left(plane[pos[i]], p[head])) ++head;
q[rear++] = plane[pos[i]];
if (abs(cross(q[rear - 1].dir() , q[rear - 2].dir())) < eps){
--rear;
if (on_left(q[rear - 1], plane[pos[i]][0])) q[rear - 1] = plane[pos[i]];
}
if (head + 1 < rear) p[rear - 2] = getLL(q[rear - 2], q[rear - 1]);
}
while (head < rear - 1 && !on_left(q[head], p[rear - 2])) --rear;
if (abs(cross(q[head].dir() , q[rear - 1].dir())) < eps){
if (on_left(q[rear - 1], q[head][0])) --rear;
else ++head;
}
if (head + 1 >= rear) return 0;
p[rear - 1] = getLL(q[rear - 1], q[head]);
n_poly = 0;
for(int i=head;i<rear;i++) poly[n_poly++] = p[i];
return 1;
}
int relation(point p,line l){ //点和向量关系 1:左侧 2:右侧 3:在线上
int c=sign(cross(p-l[0],l[1]-l[0]));
if(c<0) return 1;
else if(c>0) return 2;
else return 3;
}
line que[maxn];
int half_plane_intersection(line *L,int n){ //以逆时针方向 半平面交求多边形的核 ch表示凸包的顶点 返回顶点数 -1则表示不存在
int head=0,tail=1;
que[0]=L[0],que[1]=L[1];
for(int i=2;i<n;i++){
while(tail>head&&relation(getLL(que[tail],que[tail-1]),L[i])==2) tail--;
while(tail>head&&relation(getLL(que[head],que[head+1]),L[i])==2) head++;
que[++tail]=L[i];
}
while(tail>head&&relation(getLL(que[tail],que[tail-1]),que[head])==2) tail--;
while(tail>head&&relation(getLL(que[head],que[head+1]),que[tail])==2) head++;
for(int i=head;i<=tail;i++){
int j=(i==tail? head: i+1);
if(sign(cross(que[i][1]-que[i][0],que[j][1]-que[j][0]))<=0)
return 0;
}
return 1;
}
int T,n,m,id[400005];
vector<point> p,q;
line l[400005];int cnt=0;point s[400005];int y;
bool check(db x){
cnt=0;
for(int i=0;i<n;i++){
l[cnt++]=line(p[i]*x-q[id[i]],p[(i+1)%n]*x-q[id[i]]);
}
return half_plane_intersection(l,cnt);
}
void log(vector<point>&v){for(auto x:v)x.print();}
int main(){
scanf("%d",&T);
while (T--){
scanf("%d",&n);p.resize(n);
for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
cclock(p);n=p.size();
scanf("%d",&m);q.resize(m);
for(int i=0;i<m;i++)scanf("%lf%lf",&q[i].x,&q[i].y);
if(m==1){
printf("%.6f\n",0.0);
continue;
}
// log(q);
q=ConvexHull(q);m=q.size();
// log(q);
// log(q);
db ans = 1e18;
for(int cas=1;cas<=2;cas++){
for(int i=0;i<n;i++)p[i]=p[i]*-1;
// p=ConvexHull(p);
// log(p);
int j=0;
for(int i=0;i<n;i++){
while (sign(cross(p[(i+1)%n]-p[i],q[(j+1)%m]-q[j]))<0||sign(cross(p[(i+1)%n]-p[i],q[(j-1+m)%m]-q[j]))<0){
j=(j+1)%m;
}
id[i]=j;//这条边无限外移切到的点
// cout<<i<<' '<<j<<endl;
}
db l=0,r=1e10;
for(int i=0;i<=70;i++){
db mid = (l+r)/2;
if(check(mid))
r=mid;
else
l=mid;
}
ans = min(ans,l);
}
printf("%.11f\n",ans);
}
}