哇太妙了啊
我交了份一万多b的代码
去看了下去年jls的录播，，，毕竟这怎么做啊。。。。 然后1A了。调试信息没删不能作数的
其实很容易想到把边向里平移。。
算了我还是直接复述叭，
对每个点的圆，和所有平移R之后的线段，这两个集合求交点，也就是可行解。圆和圆，线段和线段也要求。
然后对于每个询问，先check可行解。
然后在分别check 每个圆和平移之后的线段即可。
eps开大一点。。。(1e-2)
只是单纯模拟的话板子没问题也很难出错叭
#include <bits/stdc++.h>
using namespace std;
typedef double db;
const db eps=1e-2;
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
",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;}
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(db k){
point delta=(p[1]-p[0]).turn90().unit()*k;
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));}
db closepoint(vector<point>&A,int l,int r){ // 最近点对 , 先要按照 x 坐标排序
if (r-l<=5){
db ans=1e20;
for (int i=l;i<=r;i++) for (int j=i+1;j<=r;j++) ans=min(ans,A[i].dis(A[j]));
return ans;
}
int mid=l+r>>1; db ans=min(closepoint(A,l,mid),closepoint(A,mid+1,r));
vector<point>B; for (int i=l;i<=r;i++) if (abs(A[i].x-A[mid].x)<=ans) B.push_back(A[i]);
sort(B.begin(),B.end(),[](point k1,point k2){return k1.y<k2.y;});
for (int i=0;i<B.size();i++) for (int j=i+1;j<B.size()&&B[j].y-B[i].y<ans;j++) ans=min(ans,B[i].dis(B[j]));
return ans;
}
int checkposCC(circle k1,circle k2){// 返回两个圆的公切线数量
if (cmp(k1.r,k2.r)==-1) swap(k1,k2);
db dis=k1.o.dis(k2.o); int w1=cmp(dis,k1.r+k2.r),w2=cmp(dis,k1.r-k2.r);
if (w1>0) return 4; else if (w1==0) return 3; else if (w2>0) return 2;
else if (w2==0) return 1; else return 0;
}
vector<point> getCL(circle k1,point k2,point k3){ // 沿着 k2->k3 方向给出 , 相切给出两个
point k=proj(k2,k3,k1.o); db d=k1.r*k1.r-(k-k1.o).abs2();
if (sign(d)==-1) return {};
point del=(k3-k2).unit()*sqrt(max((db)0.0,d)); return {k-del,k+del};
}
vector<point> getCC(circle k1,circle k2){// 沿圆 k1 逆时针给出 , 相切给出两个
int pd=checkposCC(k1,k2); if (pd==0||pd==4) return {};
db a=(k2.o-k1.o).abs2(),cosA=(k1.r*k1.r+a-k2.r*k2.r)/(2*k1.r*sqrt(max(a,(db)0.0)));
db b=k1.r*cosA,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
point k=(k2.o-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
return {m-del,m+del};
}
vector<point> TangentCP(circle k1,point k2){// 沿圆 k1 逆时针给出
db a=(k2-k1.o).abs(),b=k1.r*k1.r/a,c=sqrt(max((db)0.0,k1.r*k1.r-b*b));
point k=(k2-k1.o).unit(),m=k1.o+k*b,del=k.turn90()*c;
return {m-del,m+del};
}
vector<line> TangentoutCC(circle k1,circle k2){
int pd=checkposCC(k1,k2); if (pd==0) return {};
if (pd==1){point k=getCC(k1,k2)[0]; return {(line){k,k}};}
if (cmp(k1.r,k2.r)==0){
point del=(k2.o-k1.o).unit().turn90().getdel();
return {(line){k1.o-del*k1.r,k2.o-del*k2.r},(line){k1.o+del*k1.r,k2.o+del*k2.r}};
} else {
point p=(k2.o*k1.r-k1.o*k2.r)/(k1.r-k2.r);
vector<point>A=TangentCP(k1,p),B=TangentCP(k2,p);
vector<line>ans; for (int i=0;i<A.size();i++) ans.push_back((line){A[i],B[i]});
return ans;
}
}
vector<line> TangentinCC(circle k1,circle k2){
int pd=checkposCC(k1,k2); if (pd<=2) return {};
if (pd==3){point k=getCC(k1,k2)[0]; return {(line){k,k}};}
point p=(k2.o*k1.r+k1.o*k2.r)/(k1.r+k2.r);
vector<point>A=TangentCP(k1,p),B=TangentCP(k2,p);
vector<line>ans; for (int i=0;i<A.size();i++) ans.push_back((line){A[i],B[i]});
return ans;
}
int contain(vector<point>A,point q){ // 2 内部 1 边界 0 外部
int pd=0; A.push_back(A[0]);
for (int i=1;i<A.size();i++){
point u=A[i-1],v=A[i];
if (onS(u,v,q)) return 1; if (cmp(u.y,v.y)>0) swap(u,v);
if (cmp(u.y,q.y)>=0||cmp(v.y,q.y)<0) continue;
if (sign(cross(u-v,q-v))<0) pd^=1;
}
return pd<<1;
}
int T;
int n,m;db R;
point a;
vector<line> v,g;//线段
bool check(point x){
for(int i=0;i<n;i++){
if(sign(disSP(v[i][0],v[i][1],x)-R)<0)
return false;
}
return true;
}
circle c[222];
vector<point> p,s,f;
int main(){
scanf("%d",&T);
while (T--){
v.clear();g.clear();p.clear();s.clear();f.clear();
scanf("%d%d%lf",&n,&m,&R);
p.resize(n);
for(int i=0;i<n;i++){
scanf("%lf%lf",&p[i].x,&p[i].y);
c[i].o=p[i],c[i].r=R;
}
for(int i=0;i<n;i++){
v.push_back(line(p[i],p[(i+1)%n]));//原多边形
g.push_back(v[i].push(R));//移完
}
//求出来所有可行解
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(checkLL(g[i][0],g[i][1],g[j][0],g[j][1])){
point x = getLL(g[i],g[j]);
if(g[i].include(x)&&g[j].include(x)&&contain(p,x)&&check(x)){
s.push_back(x);
}
}
}
}
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
f=getCC(c[i],c[j]);
for(auto x:f){
if(contain(p,x)&&check(x)){
s.push_back(x);
}
}
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
f=getCL(c[j],g[i][0],g[i][1]);
for(auto x:f){
if(g[i].include(x)&&contain(p,x)&&check(x)){
s.push_back(x);
}
}
}
}
// for(auto x:s){
// x.print();
// }
//
while (m--){
scanf("%lf%lf",&a.x,&a.y);
if(contain(p,a)&&check(a)){
printf("0
");
continue;
}
db ans = 1e9;
for(auto x:s){
ans = min(ans,a.dis(x));
}
for(int i=0;i<n;i++){
point x = proj(g[i][0],g[i][1],a);
if(g[i].include(x)&&contain(p,x)&&check(x))
ans = min(ans,a.dis(x));
}
for(int i=0;i<n;i++){
f = getCL(c[i],c[i].o,a);
for(auto x:f){
if(contain(p,x)&&check(x)){
ans = min(ans,a.dis(x));
}
}
}
printf("%.0f
",ans);
}
}
}