Problem DescriptionAn 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 w 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.

 

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

 

Output

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

 

Sample Input

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<vector>

using namespace std;

struct Path

{

    int x,y;

    double value;

} path[];

typedef pair<int ,int> P;

vector <P> edge[];

bool cmp(Path a,Path b)

{

    return a.value<b.value;

}

int father[],n,m;;

int find(int x)

{

    if(x!=father[x])

        father[x]=find(father[x]);

    return father[x];

}

void merge(int a,int b)

{

    int x=find(a);

    int y=find(b);

    if(x!=y)

        father[x]=y;

}

long long sumv=,sumd=;

void kruscal()

{

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

        father[i]=i;

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

        edge[i].clear();

    sort(path,path+m,cmp);

    for(int i=; i<m; i++)

    {

        int x=path[i].x,y=path[i].y,value=path[i].value;

        if(find(path[i].x)!=find(path[i].y))

        {

            sumv+=path[i].value;

            merge(path[i].x,path[i].y);

            edge[x].push_back(P(y,value));

            edge[y].push_back(P(x,value));

        }

    }

}

int dfs(int x,int last)

{

    int cnt=;

    for(int i=;i<edge[x].size();i++)

    {

        int y=edge[x][i].first,value=edge[x][i].second;

        if(last!=y)

        {

            int now=dfs(y,x);

            cnt+=now;

            sumd+=1.0*now*(n-now)*value;

        }

    }

    return cnt+;

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        sumv=;

        sumd=;

        scanf("%d%d",&n,&m);

        for(int i=; i<m; i++)

        {

            scanf("%d%d%lf",&path[i].x,&path[i].y,&path[i].value);

        }

        kruscal();

        dfs(,-);

        printf("%I64d %.2lf\n",sumv,sumd*2.0/(1.0*n)/(n-1.0));

    }

    return ;

}

Sample Output

 

Author

HIT

 

Source