hdu 5723 Abandoned country

Abandoned country

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2586    Accepted Submission(s): 643

Problem Description

An abandoned country has n(n≤100000) villages which are numbered from 1 to n. Since abandoned for a long time, the roads need to be re-built. There are m(m≤1000000) roads to be re-built, the length of each road is wi(wi≤1000000). Guaranteed that any two wi are different. The roads made all the villages connected directly or indirectly before destroyed. Every road will cost the same value of its length to rebuild. The king wants to use the minimum cost to make all the villages connected with each other directly or indirectly. After the roads are re-built, the king asks a men as messenger. The king will select any two different points as starting point or the destination with the same probability. Now the king asks you to tell him the minimum cost and the minimum expectations length the messenger will walk.

 

Input

The first line contains an integer T(T≤10) which indicates the number of test cases. 

For each test case, the first line contains two integers n,m indicate the number of villages and the number of roads to be re-built. Next m lines, each line have three number i,j,wi, the length of a road connecting the village i and the village j is wi.

 

Output

output the minimum cost and minimum Expectations with two decimal places. They separated by a space.

 

Sample Input

1 4 6 1 2 1 2 3 2 3 4 3 4 1 4 1 3 5 2 4 6

 

Sample Output

6 3.33

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5733 5732 5731 5730 5729 

 

思路：

     給定一張圖，求最小生成樹，並求在圖中任取兩點，兩點間路徑代價的期望值。

     也就是求最小生成樹中：邊權*左子樹的大小*右子樹的大小的和再除以n*(n-1)；

     先求最小生成樹，並在尋找樹的過程中，保留最小生成樹上的邊，用於後續計算期望。採用dfs的方式，任意從樹上一點出發，計算該節點所在的子樹上的節點數x，並由總數減去x得到邊另一側的節點數。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef __int64 ll;
struct node
{
    ll st,en,len;
}E[1000005];
struct Edge
{
    ll en,next,len;
}edge[1000005];
ll num,head[100005],fa[100005],n,m,expect;
void init()
{
    num=0;expect=0;
    memset(head,-1,sizeof(head));
    memset(E,0,sizeof(E));
    memset(edge,0,sizeof(edge));

}
void add(ll st,ll en,ll len)
{
    edge[num].en=en;
    edge[num].len=len;
    edge[num].next=head[st];
    head[st]=num++;
}
bool cmp(node &a,node &b)
{
    return a.len<b.len;
}
ll find(ll x)
{
    if(x==fa[x]) return x;
    return fa[x]=find(fa[x]);
}
ll kruskal()
{
    ll cnt=0,ans=0;
      for(ll i=0;i<=n;i++)
        fa[i]=i;
    for(ll i=0;i<m;i++)
    {
        ll fx=find(E[i].st);
        ll fy=find(E[i].en);
        if(fx!=fy)
        {
            ans+=E[i].len;
            fa[fx]=fy;
            cnt++;
            add(E[i].st,E[i].en,E[i].len);
            add(E[i].en,E[i].st,E[i].len);
            if(cnt==n-1) break;
        }
    }
    return ans;
}
ll dfs(ll x,ll fa)
{
    ll tmp,res=1;
    for(ll i=head[x];i!=-1;i=edge[i].next)
    {
        if(edge[i].en!=fa)
        {
            tmp=dfs(edge[i].en,x);
            expect+=1.0*(n-tmp)*tmp*edge[i].len;
            res+=tmp;
        }
    }
    return res;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%I64d%I64d",&n,&m);
        for(ll i=0;i<m;i++)
        {
            scanf("%I64d%I64d%I64d",&E[i].st,&E[i].en,&E[i].len);
        }
        sort(E,E+m,cmp);
        ll ans=kruskal();
        dfs(1,-1);
        double ans1=2.0*expect/((n-1)*n);
        printf("%I64d %.2lf\n",ans,ans1);
    }
    return 0;
}