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Arctic Network
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 1 Sample Output 212.13 Source |
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题意：有S颗卫星和P个哨所，有卫星的两个哨所之间可以任意通信；否则，一个哨所只能和距离它小于等于D的哨所通信。给出卫星的数量和P个哨所的坐标，求D的最小值。
思路：找出一棵最小生成树，用卫星的代替长的边，剩下的最长的边就是答案
#include <iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
#define LL long long struct node
{
int u,v;
double w;
} p[10000005];
int n,cnt,x,pre[10006];
bool cmp(node a,node b)
{
return a.w<b.w;
}
void init()
{
for(int i=0; i<10005; i++)
pre[i]=i;
} int fin(int x)
{
return pre[x]==x?x:pre[x]=fin(pre[x]);
} void kruskal()
{
sort(p,p+cnt,cmp);
init();
double cost=0;
int ans=0;
for(int i=0; i<cnt; i++)
{
int a=fin(p[i].u);
int b=fin(p[i].v);
if(a!=b)
{
pre[a]=b;
cost=max(cost,p[i].w);
ans++;
}
if(ans==n-x)
{
break;
}
}
printf("%.2f\n",cost);
} int main()
{
int T;
double a[10006],b[10006];
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&x,&n);
for(int i=0;i<n;i++)
scanf("%lf%lf",&a[i],&b[i]);
cnt=0;
for(int i=0; i<n; i++)
for(int j=i+1; j<n; j++)
{
p[cnt].u=i,p[cnt].v=j;
p[cnt++].w=sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
}
kruskal();
}
return 0;
}