WOJ 656 最小生成树 Prim

656. Wifi Relay

Time Limit: 2 second

The ACM/ICPC team has bought $n$ wifi APs. And the APs have a special feature called Wifi Relay which means if two APs' signal can cover each other, then they can work like a single wifi AP.

Each AP can cover a round area, the radius of which is $r$.

In a programming contest, GSS set up $n$ APs, and the $i$-th AP is placed at (xi,yi)(x_i, y_i)(x​i​​,y​i​​). Then GSS use the Wifi Relay to let all the APs work like a single AP that every one can access the system without being interrupted.

Then GSS has to set all the radius each APs can cover to $r$. To save pover, GSS has to get the minimal $r$ he should set, can you help him to do that?

Input

The first line of input is an integer $n$. (2≤n≤1042 \le n \le 10^42≤n≤10​4​​)

The following $n$ lines, the $i$-th one of them contains two integers (xi,yi)(x_i, y_i)(x​i​​,y​i​​), the position GSS place $i$-th AP. (0≤xi,yi≤1080\le x_i, y_i \le 10^80≤x​i​​,y​i​​≤10​8​​)

Output

Output one integer, the minimal $r$ GSS should set. Your answer will be considered as correct if the relative or absolute error is less than10−910^{-9}10​−9​​

Examples

Input 1

30 010 1010 0

Output 1

5.0000000000

Source

第六届华中区程序设计邀请赛暨武汉大学第十五届校赛

        如此水题当时竟然没有做出来？罪过啊。

        其实一切都是因为没有信心写。看着O(N^2)的复杂度，根本不敢下手……其实据说O(N^2logN)的算法都可以解……

        具体说说吧，最小生成树中最长的边即为本题的解。下面给出自己简短的证明。假设存在一个解x，比最小生成树的最长边y要短，那么就说明能够以长度小于等于x的边构成一颗生成树。根据kruskal最小生成树的求解法，如果用长度小于等于x的边已经能够构成最小生成树了，那么边y>x就不是最小生成树中的边，与题设矛盾，所以这个最长边y就是本题的解。

        然后考虑到本题是一个稠密图，所以用Prim求解最小生成树更为快速。具体代码：

#include<bits/stdc++.h>#define INF 1e17using namespace std;struct node{int x,y;} wifi[10010];inline long long sqr(long long x) {return x*x;}inline long long dist(node a,node b){return sqr(a.x-b.x)+sqr(a.y-b.y);}long long d[10010],n;bool v[10110];int main(){cin>>n;for(int i=1;i<=n;i++){scanf("%d%d",&wifi[i].x,&wifi[i].y);d[i]=INF;}v[1]=1; int last=1;long long ans=0;for(int i=1;i<n;i++){int k=0;long long cur=INF;for(int j=1;j<=n;j++){if (v[j]) continue;d[j]=min(d[j],dist(wifi[j],wifi[last]));if (d[j]<cur) k=j,cur=d[j];}ans=max(ans,d[k]);v[k]=1; last=k;}cout << fixed << setprecision(12) << sqrt(ans) / 2 << endl;return 0;}