[poj2349]Arctic Network(最小生成树+贪心)
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 17758 | Accepted: 5646 |
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<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef struct{
int frm,to;
double dis;
}edge;
typedef struct{
int x,y;
}point;
edge gra[];
point poi[];
int fa[],num;
int fnd(int x){
return fa[x]==x?x:fnd(fa[x]);
}
int uni(int x,int y){
int fx=fnd(x);
int fy=fnd(y);
fa[fy]=fx;
return ;
}
int cmp(const edge &a,const edge &b){
return a.dis<b.dis;
}
int add(int frm,int to,double dis){
gra[++num].frm=frm;
gra[num].to=to;
gra[num].dis=dis;
return ;
}
double kru(int k){
int cnt=;
sort(gra+,gra+num+,cmp);
for(int i=;i<=num;i++){
int fx=fnd(gra[i].frm);
int fy=fnd(gra[i].to);
if(fx!=fy){
cnt++;
if(cnt==k){
return gra[i].dis;
}
uni(fx,fy);
}
}
return 0.0;
}
int main(){
int s,p,t;
scanf("%d",&t);
while(t--){
num=;
scanf("%d %d",&s,&p);
if(s==)s=;
for(int i=;i<=p;i++)scanf("%d %d",&poi[i].x,&poi[i].y);
for(int i=;i<=p;i++)fa[i]=i;
for(int i=;i<=p;i++){
for(int j=i+;j<=p;j++){
double dis=sqrt((double)((poi[i].x-poi[j].x)*(poi[i].x-poi[j].x)+(poi[i].y-poi[j].y)*(poi[i].y-poi[j].y)));
add(i,j,dis);
add(j,i,dis);
}
}
printf("%.2lf\n",kru(p-s));
}
return ;
}
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