题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5723

Abandoned country

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2969    Accepted Submission(s): 725

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

题意：

给出n个点和m条边，每条边的权值都不相同，求使得n个点连通而边的权值之和最小，以及随机选2个点作为起点和终点的路径长度的期望

题解：

最小权值直接求最小生成树就好

m条边权值各不相同，所以每个权值对应一个期望，求出最小生成树的时候，期望也一定是最小的

期望是等于每条边的贡献之和除以总路径。

每条边两边的点连通都要经过这条边，所以这条边的贡献就是s（其中一边的点的数）*（n-s）*w（权值）；

求每条边的一边的点，可以利用dfs

源代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=100005;
int n,m;
struct node
{
    int x,y,w;
    node(int xx,int yy,int ww):x(xx),y(yy),w(ww) {}
};
vector<node> a;
struct Node
{
    int y,w;
    Node(int yy,int ww):y(yy),w(ww) {}
};
vector<Node> b[maxn];
int f[maxn];
bool cmp(const node &i,const node &j)
{
    return i.w<j.w;
}
int finds(int x)
{return f[x] == x?x:f[x] = finds(f[x]);}
long long Kruskal()
{
    long long ans=0;
    for(int i=1; i<=n; i++) f[i]=i;
    sort(a.begin(),a.end(),cmp);
    for(int i=0; i<a.size(); i++)
    {
        int x=finds(a[i].x);
        int y=finds(a[i].y);
        if(x==y)continue;
        f[x]=y;
        ans+=a[i].w;
        b[a[i].x].push_back(Node(a[i].y,a[i].w));
        b[a[i].y].push_back(Node(a[i].x,a[i].w));
    }
    return ans;
}
int s[maxn];
double d[maxn];
void dfs(int root,int father)
{
    s[root]=1;
    for(int i=0; i<b[root].size(); i++)
    {
        int son=b[root][i].y;
        int l=b[root][i].w;
        if(son==father) continue;
        dfs(son,root);
        s[root]+=s[son];
        d[root]+=d[son]+((double)s[son]*(n-s[son]))*l;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        a.clear();
        for(int i=0; i<m; i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            a.push_back(node(x,y,w));
        }
        memset(s,0,sizeof(s));
        memset(d,0,sizeof(d));
        for(int i=1; i<=n; i++)
            b[i].clear();
        long long ans=Kruskal();
        dfs(1,0);
        long long g=(long long)n*(n-1)/2;
        printf("%I64d %.2lf\n",ans,d[1]/(double)g);
    }
    return 0;
}