Codeforces Round #250 (Div. 2)D

给你一张无向图,每个点有一个权值,对于一条从l到r 的边权值是l到r路径上最小的点的权值,(多条路取最大的权值),然后求每两个点之间的权值和/点对数

题解:并查集维护,先从点大的边排序,然后依次加边,这样每次加进来的保证是当前最大 的,然后每次合并都要加上两端的最小值*两端的size,类似与每一个最小值算贡献

#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pii pair<int,int>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)

using namespace std;

const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f3f;

ll a[N],father[N],sz[N];
struct edge{
    ll from,to,c;
}e[N];
bool cmp(edge x,edge y)
{
    return x.c > y.c;
}
ll Find(ll x)
{
    return x==father[x]?x:father[x]=Find(father[x]);
}
int main()
{
    ll n,m;
    scanf("%lld%lld",&n,&m);
    for(ll i=;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        father[i]=i;
        sz[i]=;
    }
    for(int i=;i<m;i++)
    {
        scanf("%lld%lld",&e[i].from,&e[i].to);
        e[i].c = min(a[e[i].from], a[e[i].to]);
    }
    sort(e,e+m,cmp);
    ll ans=;
    for(int i=;i<m;i++)
    {
        ll x=e[i].from,y=e[i].to;
//        cout<<x<<" "<<y<<" "<<sz[x]<<" "<<sz[y]<<endl;
        ll fx=Find(x),fy=Find(y);
        if(fx!=fy)
        {
            ans+=min(a[x],a[y])*sz[fx]*sz[fy];
            father[fx]=fy;
            sz[fy]+=sz[fx];
        }
    }
    ll te=(n-)*n/;
    printf("%.12f\n",(double)ans/te);
    return ;
}
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