POJ 2349 Arctic Network （最小產生樹）

標籤：cos   fence   direct   contain   最大的   mmu   return   determine   int   

Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel. 
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts. 

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

12 40 1000 3000 600150 750

Sample Output

212.13
題意：
　　可以這麼理解：有S個衛星可以免費通訊，相當於有S-1條權值為0的邊，然後剩下求出剩下權值最大的一條邊。
題解：
　　　當然是最小產生樹裡最大的m-1條邊權值設為0，然後記錄剩下最大的那條邊的權值就好了。

#include<iostream>#include<cmath>#include<algorithm>using namespace std;const int maxn=505;int par[maxn];int n,m;int cas;struct edge{    int u,v;    double cost;}es[maxn*maxn];struct node{    int x,y;}p[maxn];double dis(node a,node b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool cmp(edge a,edge b){    return a.cost<b.cost;}void init(){    for(int i=1;i<=n;i++)        par[i]=i;}int find(int x){    return x==par[x]?x:par[x]=find(par[x]);}void unite(int x,int y){    x=find(x);    y=find(y);    if(x!=y)        par[x]=y;}bool same(int x,int y){    return find(x)==find(y);}double kruskal(){    init();    sort(es,es+cas,cmp);    int cnt=0;    for(int i=0;i<cas;i++)    {        edge e=es[i];        if(!same(e.u,e.v))        {            unite(e.u,e.v);            if(++cnt==n-m)                return e.cost;        }    }    return 0;}int main(){        int t;    cin>>t;    while(t--)    {        cin>>m>>n;        for(int i=0;i<n;i++)            cin>>p[i].x>>p[i].y;        cas=0;        for(int i=0;i<n;i++)            for(int j=i+1;j<n;j++)            {                double d=dis(p[i],p[j]);                es[cas].u=i+1,es[cas].v=j+1;                es[cas].cost=d;                cas++;            }        printf("%.2lf\n",kruskal());    }    return 0;}

 

POJ 2349 Arctic Network （最小產生樹）