Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 10091 | Accepted: 3342 |
Description
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
Output
Sample Input
1 2 4 0 100 0 300 0 600 150 750
Sample Output
212.13
Source
#include <stdio.h>
#include <math.h>
#include <algorithm>
#define MAX 505
using namespace std;
struct KK{
int u;
int v;
double w;
}a[MAX*MAX/2];
struct PP{
double x;
double y;
}pp[MAX];
int father[MAX];
double d(double i,double j,double m,double n){
return sqrt((m-i)*(m-i)+(n-j)*(n-j));
}
bool operator <(KK i,KK j){
return i.w<j.w;
}
int findfather(int x){
if(x!=father[x])
return father[x]=findfather(father[x]);
return x;
}
int main()
{
int n;
scanf("%d",&n);
while(n--){
int s,p;
scanf("%d%d",&s,&p);
for(int i=1;i<=p;i++){
scanf("%lf%lf",&pp[i].x,&pp[i].y);
}
// for(int i=1;i<=p;i++)
// printf("%.2f %.2f \n",pp[i].x,pp[i].y);
int t=1;
for(int i=1;i<p;i++){
for(int j=i+1;j<=p;j++){
a[t].u=i;
a[t].v=j;
a[t].w=d(pp[i].x,pp[i].y,pp[j].x,pp[j].y);
t++;
// printf("%.2f\n",a[t-1].w);
}
}
sort(a+1,a+t);
// for(int i=1;i<=p*(p-1)/2;i++)
// printf("%d %d %.2lf\n",a[i].u,a[i].v,a[i].w);
for(int i=1;i<=p;i++)
father[i]=i;
double result[MAX];
int count=1;
for(int i=1;i<t;i++){
int x=findfather(a[i].u);
int y=findfather(a[i].v);
if(x!=y){
result[count++]=a[i].w;
father[x]=y;
// printf("%.2lf%d\n",result[i]);
}
}
printf("%.2lf\n",result[p-s]);
}
return 0;
}