[2016-04-13] [HDU] [1875] [Smooth works re-continued]

• Time: 2016-04-13-23:58:24
• Title Number: [2016-04-13][hdu][1875][smooth works re-continued]
• The smallest spanning tree in the specified range is the right edge
• Analysis: Kruskal The merger of the time to judge, and finally determine whether the combined edge number is n-1

  
 

   
  

  
#include<cstdio>
   
  

  
#include<algorithm>
   
  

  
#include<cmath>
   
  

  
using namespace std;
   
  

  
const int maxn = 100 + 10;
   
  

  
struct Point{
   
  

  
 int x,y;
   
  

  
 Point(int a=0,int b=0):x(a),y(b){}
   
  

  
}p[maxn];
   
  

  
struct Edge{
   
  

  
  int    v    C  ;   
   
  

  
  edge   ( int   a  =  0   int   b  =  0   int   cc  =  0  ):  u   ( a   v   ( b   c   ( cc  ) {}   
   
  

  
 bool operator < (const Edge & a)const{
   
  

  
 return c < a.c;
   
  

  
 }
   
  

  
}e[maxn * maxn];
   
  

  
int fa[maxn];
   
  

  
int fnd(int x){
   
  

  
 return x == fa[x]?x:fa[x] = fnd(fa[x]);
   
  

  
}
   
  

  
void ini(int n){
   
  

  
 for(int i = 0 ; i <= n ; ++i){
   
  

  
 fa[i] = i;
   
  

  
 }
   
  

  
}
   
  

  
int dis(int a,int b){
   
  
 return (P[a].x-P[b].x)*(P[a].x-P[b].x) + (P[a].y-P[b].y) * (P[a].y-P[b].y);
   
  

  
}
   
  

  

   
  

  
int main(){
   
  

  
 int t;
   
  

  
 scanf("%d",&t);
   
  

  
 while(t--){
   
  

  
 int n;
   
  

  
 scanf("%d",&n);
   
  

  
 ini(n);
   
  

  
 for(int i = 0 ; i < n ; ++i){
   
  

  
 scanf("%d%d",&p[i].x,&p[i].y); 
   
  

  
 } 
   
  

  
 int cnt = 0;
   
  

  
  for   ( int   i  =    0    I  <   n     ++  i  ) {  
   
  

  
  for   ( int   J  =    0    ;   J  <   n     ++  j  ) {  
   
  

  
 e[cnt++] = Edge(i,j,dis(i,j));
   
  

  
 }
   
  

  
 } 
   
  

  
 sort(e,e+cnt);
   
  

  
 int k = 0;double ans = 0;
   
  

  
 for(int i = 0 ; i < cnt ; ++i){
   
  

  
 int f1 = fnd(e[i].u);
   
  

  
  int   F2  =   fnd   (  E  [ i  .  v    
   
  

  
  if   ( f1  !=   F2   &&   e  [ i   c  >=    100    &&   e  [ i  .  c  <=    1000  *  1000  Span class= "pun") {  
   
  

  
 fa[f1] = f2;
   
  

  
 ans += sqrt(double( e[i].c));
   
  

  
 ++k;
   
  

  

   
  

  
 }
   
  

  
 if(k == n - 1) break;
   
  

  
 } 
   
  

  
 if(k == n - 1)
   
  

  
 printf("%.1lf\n",ans * 100);
   
  

  
 else puts("oh!");
   
  

  

   
  

  
 }
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-13] [HDU] [1875] [Smooth works re-continued]