CF437D(The Child and Zoo)最小生成树-白红宇

CF437D(The Child and Zoo)最小生成树

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发布时间：2019-06-27

本文共 2724 字，大约阅读时间需要 9 分钟。

题目：

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 $\text{[math]}$ .

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

       Sample test(s) 
         input 
4 310 20 30 401 32 34 3
         output 
16.666667
         input 
3 310 20 301 22 33 1
         output 
13.333333
         input 
7 840 20 10 30 20 50 401 22 33 44 55 66 71 45 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 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             #include 
              
               #include 
               
                #include 
                
                 //freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8const 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

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