Problem:
Score : 600 points

Problem Statement

On the xy-plane, Snuke is going to travel from the point (x_s, y_s) to the point (x_t, y_t).
He can move in arbitrary directions with speed 1.
Here, we will consider him as a point without size.

There are N circular barriers deployed on the plane.
The center and the radius of the i-th barrier are (x_i, y_i) and r_i, respectively.
The barriers may overlap or contain each other.

A point on the plane is exposed to cosmic rays if the point is not within any of the barriers.

Snuke wants to avoid exposure to cosmic rays as much as possible during the travel.
Find the minimum possible duration of time he is exposed to cosmic rays during the travel.

Constraints

All input values are integers.

-10^9 ≤ x_s, y_s, x_t, y_t ≤ 10^9

(x_s, y_s) ≠ (x_t, y_t)

1≤N≤1,000

-10^9 ≤ x_i, y_i ≤ 10^9

1 ≤ r_i ≤ 10^9

Input

The input is given from Standard Input in the following format:

x_s y_s x_t y_t
N
x_1 y_1 r_1
x_2 y_2 r_2
:
x_N y_N r_N

Output

Print the minimum possible duration of time Snuke is exposed to cosmic rays during the travel.
The output is considered correct if the absolute or relative error is at most 10^{-9}.

Sample Input 1

-2 -2 2 2
1
0 0 1

Sample Output 1

3.6568542495

An optimal route is as follows:

Sample Input 2

-2 0 2 0
2
-1 0 2
1 0 2

Sample Output 2

0.0000000000

An optimal route is as follows:

Sample Input 3

4 -2 -2 4
3
0 0 2
4 0 1
0 4 1

Sample Output 3

4.0000000000

An optimal route is as follows:

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h> // ver2.7.1
#define int long long
#define endl "\n" // fflush(stdout);
#define ALL(v) (v).begin(),(v).end()
#define Vi vector<int>
#define VVi vector<Vi>
#define VVVi vector<VVi>
#define Vm vector<mint>
#define Vs vector<string>
#define Vd vector<double>
#define Vc vector<char>
#define Pii pair<int,int>
#define Pdd pair<double,double>
#define VPii vector<Pii>
#define Tiii tuple<int,int,int>
#define VTiii vector<Tiii>
#define PQi priority_queue<int>
#define PQir priority_queue<int,vector<int>,greater<int>>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define itos to_string
#define stoi stoll
#define FI first
#define SE second
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define cyes cout<<"yes"<<endl
#define cno cout<<"no"<<endl
#define sortr(v) sort(v,greater<>())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repeq(i,a,b) for(int i=a;i<=b;i++)
#define repreq(i,a,b) for(int i=a;i>=b;i--)
#define dem(a,b) ((a+b-1)/(b))
#define INF 3000000000000000000 // 3.0*10^18
#define MAX LLONG_MAX
#define PI acos(-1.0L)
//int MOD = 998244353; // case
const int MOD = 1000000007; // 10^9 + 7
const double EPS = 1e-4;

using namespace std;

struct mint{int x;mint(int x=0):x((x%MOD+MOD)%MOD){}
mint operator-()const{return mint(-x);}
mint& operator+=(const mint a){if((x+=a.x)>=MOD)x-=MOD;return*this;}
mint& operator-=(const mint a){if((x+=MOD-a.x)>=MOD)x-=MOD;return*this;}
mint& operator*=(const mint a){(x*=a.x)%=MOD;return*this;}
mint operator+(const mint a)const{return mint(*this)+=a;}
mint operator-(const mint a)const{return mint(*this)-=a;}
mint operator*(const mint a)const{return mint(*this)*=a;}
mint pow(int t)const{if(!t)return 1;mint a=pow(t>>1);a*=a;if(t&1)a*=*this;return a;}
mint inv()const{return pow(MOD-2);}
mint& operator/=(const mint a){return*this*=a.inv();}
mint operator/(const mint a)const{return mint(*this)/=a;}};
istream& operator>>(istream& is, const mint& a) {return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) {return os << a.x;}

/* debug */
template <typename T> // vector
ostream& operator<<(ostream& os,const vector<T> &V){int N=V.size();
if(N==0){os<<'.';return os;}rep(i,0,N-1){os<<V[i]<<' ';}os<<V[N-1];return os;}
template <typename T> // vector vector
ostream& operator<<(ostream& os,const vector<vector<T>> &V){
int N=V.size();rep(i,0,N-1)cout<<V[i]<<endl;cout<<V[N-1];return os;}
template <typename T,typename S> // pair
ostream& operator<<(ostream& os, pair<T,S> const&P){os<<P.FI<<' '<<P.SE;return os;}
template <typename T> // set
ostream& operator<<(ostream& os,set<T> &S){auto it=S.begin();while(it!=S.end()){
os<<*it;if(next(it,1)!=S.end())os<<' ';it++;};return os;}
template <typename T> // deque
ostream& operator<<(ostream& os, deque<T> &q){for(auto it=q.begin();it<q.end();it++){
os<<*it;if(it+1!=q.end())os<<" ";}return os;}

/* useful */
template<typename T>void Vin(vector<T> &v){int n=v.size();rep(i,0,n)cin>>v[i];}
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
int mypow(int x, int n, int m){
if(n==0)return 1;if(n%2==0)return mypow(x*x%m,n/2,m);else return x*mypow(x,n-1,m)%m;}
int scomb(int n, int r){if(r<0||r>n)return 0;if((n-r)<r)r=n-r; // nCr
int a=1;for(int i=n;i>n-r;--i){a=a*i;}for(int i=1;i<r+1;++i){a=a/i;}return a;}
Vi vis(Vi &v){Vi S(v.size()+1);rep(i,1,S.size())S[i]+=v[i-1]+S[i-1];return S;}
int digitsum(int n){int ret=0; while(n>0){ret+=n%10;n/=10;}return ret;}
int digit(int k,int i){string s = itos(k);return s[s.size()-i]-'0';} // i桁目の数字
template<typename T>void press(T &v){v.erase(unique(ALL(v)),v.end());} // 圧縮
Vi zip(Vi b){int Z=b.size(); // 座標圧縮
Pii p[Z+10];int a=b.size();Vi l(a);for(int i=0;i<a;i++)
p[i]=mp(b[i],i);sort(p,p+a);int w=0;for(int i=0;i<a;i++)
{if(i&&p[i].first!=p[i-1].first)w++;l[p[i].second]=w;}return l;}
Vi nibe2V(){Vi a(60);int q = 1;rep(i,0,60){a[i]=q;q*=2;}return a;}
//Vi nibe = nibe2V(); // 2^n
int modiv(int a,int b){return a*mypow(b,MOD-2,MOD)%MOD;} // a÷b(%MOD)
int SMALLER(Vi &a,int x){return lower_bound(a.begin(),a.end(),x)-a.begin();}
int orSMALLER(Vi &a,int x){return upper_bound(a.begin(),a.end(),x)-a.begin();}
int BIGGER(Vi &a,int x){return a.size()-orSMALLER(a,x);}
int orBIGGER(Vi &a,int x){return a.size()-SMALLER(a,x);}
int COUNT(Vi &a,int x) {return upper_bound(ALL(a),x)-lower_bound(ALL(a),x);}
int maxind(Vi &a){return max_element(ALL(a))-a.begin();}
int minind(Vi &a){return min_element(ALL(a))-a.begin();}
Vi stpowV(){Vi a(1000005);a[0]=1;repeq(i,1,1000004)a[i]=a[i-1]*i%MOD;return a;}
//Vi stpow = stpowV(); // 階乗配列(%MOD)

/****************************** START ******************************/

struct graph_p {
  
  int N;
  vector<vector<pair<int,double>>> gr;
  vector<double> dist;
  vector<int> prever;
  
  graph_p(int n){
    init(n);
  }
  
  void init(int n){
    N = n;
    gr.resize(n);
    dist.resize(n,INF);
  }
  
  void add_edge(int a,int b,double c){
    gr[a].pb(mp(b,c));
  }
  
  void dijkstra(int s){
    int n = gr.size();
    dist = vector<double>(n);
    prever = vector<int>(n,-1);
    rep(i,0,n) dist[i]=INF;
    dist[s] = 0;
    
    priority_queue<tuple<double,int,int>,vector<tuple<double,int,int>>,
    greater<tuple<double,int,int>>> q;
    q.push(make_tuple(0,s,-1));
    while(!q.empty()){
      double nowdist,now,past;
      tie(nowdist,now,past)=q.top();
      q.pop();
      if(nowdist>dist[now]) continue;
      dist[now]=nowdist;
      rep(i,0,gr[now].size()){
        double next=gr[now][i].first,cost=gr[now][i].second;
        if(next!=past && nowdist+cost<dist[next]){
          prever[next] = now; // テスト
          q.push(make_tuple(nowdist+cost,next,now));
          dist[next]=nowdist+cost;
        }
      }
    }
    return;
  }

  vector<int> get_path(int t){ //頂点tへの最短路
    if(dist[t] >= INF) return {-1};
    vector<int> path;
    for(;t!=-1;t=prever[t]){
      path.push_back(t);
    }
    reverse(path.begin(), path.end());
    return path;
  }
  
};

double dd(Pdd A,Pdd B){ // 2点間の距離
   return sqrt((A.FI-B.FI)*(A.FI-B.FI)+(A.SE-B.SE)*(A.SE-B.SE));
 }

void sol(){
  int a,b,c,d;cin >> a >> b >> c >> d;
  int n;cin >> n;
  n += 2;
  Vi x(n),y(n),r(n);
  rep(i,0,n-2){
    cin >> x[i+1] >> y[i+1] >> r[i+1];
  }
  x[0] = a;
  y[0] = b;
  x[n-1] = c;
  y[n-1] = d;
  graph_p gr(n);
  
  rep(i,0,n)rep(j,0,n){
    if(i==j) continue;
    double now = dd(mp(x[i],y[i]),mp(x[j],y[j]));
    //cout << now << endl;
    gr.add_edge(i,j,max(0.0,now-r[i]-r[j]));
  }
  gr.dijkstra(0);
  cout << gr.dist[n-1] << endl;
}

signed main() {
  cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(12);
  int q = 1;
  //cin >> q;
  rep(i,0,q) sol();
  return 0;
}