1 //2017-08-27
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <iomanip>
  8 #include <cmath>
  9 
 10 using namespace std;
 11 
 12 const int N = 5010;
 13 const int M = N*N;
 14 const double EPS = 1e-6;
 15 int head[N], rhead[N], tot, rtot;
 16 struct Edge{
 17     int to, next;
 18 }edge[M], redge[M];
 19 
 20 void init(){
 21     tot = 0;
 22     rtot = 0;
 23     memset(head, -1, sizeof(head));
 24     memset(rhead, -1, sizeof(rhead));
 25 }
 26 
 27 void add_edge(int u, int v){
 28     edge[tot].to = v;
 29     edge[tot].next = head[u];
 30     head[u] = tot++;
 31 
 32     redge[rtot].to = u;
 33     redge[rtot].next = rhead[v];
 34     rhead[v] = rtot++;
 35 }
 36 
 37 vector<int> vs;//后序遍历顺序的顶点列表
 38 bool vis[N];
 39 int cmp[N];//所属强连通分量的拓扑序
 40 
 41 //input: u 顶点
 42 //output: vs 后序遍历顺序的顶点列表
 43 void dfs(int u){
 44     vis[u] = true;
 45     for(int i = head[u]; i != -1; i = edge[i].next){
 46         int v = edge[i].to;
 47         if(!vis[v])
 48           dfs(v);
 49     }
 50     vs.push_back(u);
 51 }
 52 
 53 //input: u 顶点编号; k 拓扑序号
 54 //output: cmp[] 强连通分量拓扑序
 55 void rdfs(int u, int k){
 56     vis[u] = true;
 57     cmp[u] = k;
 58     for(int i = rhead[u]; i != -1; i = redge[i].next){
 59         int v = redge[i].to;
 60         if(!vis[v])
 61           rdfs(v, k);
 62     }
 63 }
 64 
 65 //Strongly Connected Component 强连通分量
 66 //input: n 顶点个数
 67 //output: k 强连通分量数;
 68 int scc(int n){
 69     memset(vis, 0, sizeof(vis));
 70     vs.clear();
 71     for(int u = 0; u < n; u++)
 72       if(!vis[u])
 73         dfs(u);
 74     int k = 0;
 75     memset(vis, 0, sizeof(vis));
 76     for(int i = vs.size()-1; i >= 0; i--)
 77       if(!vis[vs[i]])
 78         rdfs(vs[i], k++);
 79     return k;
 80 }
 81 
 82 int n;
 83 struct Point{
 84     int x, y;
 85 }point[N];
 86 
 87 //input: 两个点
 88 //output: 两点间距离
 89 double distance(Point a, Point b){
 90     return sqrt((double)(a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
 91 }
 92 
 93 //input：radius 半径
 94 //output：true 通过选取某些点可以得到radius的分数，false 无法得到radius的分数
 95 bool check(double radius){
 96     init();
 97     for(int i = 0; i < 2*n; i++){
 98         for(int j = i+1; j < 2*n; j++){
 99             if((i^1) == j)continue;
100             if(distance(point[i], point[j]) < 2*radius){//i与j存在矛盾
101                 add_edge(i^1, j);// NOT i -> j
102                 add_edge(j^1, i);// NOT j -> i
103             }
104         }
105     }
106     scc(2*n);
107     for(int i = 0; i < 2*n; i += 2){
108         if(cmp[i] == cmp[i^1])
109               return false;
110     }
111     return true;
112 }
113 
114 int main()
115 {
116     std::ios::sync_with_stdio(false);
117     //freopen("inputC.txt", "r", stdin);
118     while(cin>>n){
119         for(int i = 0; i < n; i++){
120             cin>>point[2*i].x>>point[2*i].y>>point[2*i+1].x>>point[2*i+1].y;
121         }
122         double l = 0.0, r = 40000.0, mid, ans = 0;
123         while(r-l > EPS){
124             mid = (l+r)/2;
125             if(check(mid)){
126                 ans = mid;
127                 l = mid;
128             }else
129                   r = mid;
130         }
131         cout.setf(ios::fixed);
132         cout<<setprecision(2)<<ans<<endl;
133     }
134 
135     return 0;
136 }