CF437D(The Child and Zoo)最小生成树

题目：

D. The Child and Zoo

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Of course our child likes walking in a zoo. The zoo has n areas, that are numbered from 1 to n. The i-th area contains ai animals in it. Also there are m roads in the zoo, and each road connects two distinct areas. Naturally the zoo is connected, so you can reach any area of the zoo from any other area using the roads.

Our child is very smart. Imagine the child want to go from area p to area q. Firstly he considers all the simple routes from p to q. For each route the child writes down the number, that is equal to the minimum number of animals among the route areas. Let's denote the largest of the written numbers as f(p, q). Finally, the child chooses one of the routes for which he writes down the value f(p, q).

After the child has visited the zoo, he thinks about the question: what is the average value of f(p, q) for all pairs p, q (p ≠ q)? Can you answer his question?

Input

The first line contains two integers n and m (2 ≤ n ≤ 105; 0 ≤ m ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each line contains two integers xi and yi (1 ≤ xi, yi ≤ n; xi ≠ yi), denoting the road between areas xi and yi.

All roads are bidirectional, each pair of areas is connected by at most one road.

Output

Output a real number — the value of .

The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 4.

Sample test(s)

input

4 3
10 20 30 40
1 3
2 3
4 3

output

16.666667

input

3 3
10 20 30
1 2
2 3
3 1

output

13.333333

input

7 8
40 20 10 30 20 50 40
1 2
2 3
3 4
4 5
5 6
6 7
1 4
5 7

output

18.571429

解法：

定义每一个边的权值等于两个点中较小一个点的点权。

然后并查集求最大生成树。每次从加边权最大的边。然后以此最小的组合就是两头全部点的相互组合。 

代码：

/******************************************************
* author:xiefubao
*******************************************************/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string.h>
//freopen ("in.txt" , "r" , stdin);
using namespace std;


#define eps 1e-8
const double pi=acos(-1.0);
typedef long long LL;
const int Max=101000;
const int INF=1000000007 ;


int parent[Max];
LL count1[Max];
int num[Max];
int getparent(int t)
{
    if(t==parent[t])
        return t;
    return parent[t]=getparent(parent[t]);
}
int n,m;
struct point
{
    int u,v;
    LL value;
} points[Max];
bool operator<(const point& a,const point& b)
{
    return a.value>b.value;
}
int main()
{
  while(scanf("%d%d",&n,&m)==2)
  {
      for(int i=1;i<=n;i++)
      scanf("%d",num+i);
      for(int i=1;i<=n;i++)
      {
          parent[i]=i;
          count1[i]=1;
      }
      for(int i=0;i<m;i++)
      {
          scanf("%d%d",&points[i].u,&points[i].v);
          points[i].value=min(num[points[i].u],num[points[i].v]);
      }
      sort(points,points+m);
      double ans=0;
      for(int i=0;i<m;i++)
      {
          int t1=getparent(points[i].u);
          int t2=getparent(points[i].v);
          if(t1==t2)
            continue;
          parent[t2]=t1;
          ans+=count1[t1]*count1[t2]*points[i].value;
          count1[t1]+=count1[t2];
      }
      LL N=n;
      printf("%.6f\n",ans/(N*(N-1))*2.0);
  }
   return 0;
}