Abandoned country

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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

Sample Output

6 3.33

分析：Kruscal算法求最短路并建树，然后每条边对答案的贡献是分别两个端点及其外部的点的个数

　　　的乘积乘上边长/总可能情况数，dfs回溯求解点的个数；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=1e6+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m,p[maxn],vis[maxn],cnt;
double ans[2];
vector<pii>edge[maxn];
struct node
{
    int a,b,l;
}q[maxn];
bool cmp(const node&x,const node&y)
{
    return x.l<y.l;
}
int all(int x){return p[x]==x?x:p[x]=all(p[x]);}
int dfs(int now)
{
    int son=1,son_son;
    vis[now]=1;
    for(auto x:edge[now])
    {
        if(!vis[x.fi]){
            son_son=dfs(x.fi);
            son+=son_son;
            ans[1]+=1.0*son_son*(n-son_son)*x.se;
        }
    }
    return son;
}
int main()
{
    int i,j,k,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        rep(i,1,n)p[i]=i,edge[i].clear();
        memset(vis,0,sizeof(vis));
        ans[0]=ans[1]=0;cnt=0;
        rep(i,0,m-1)scanf("%d%d%d",&q[i].a,&q[i].b,&q[i].l);
        sort(q,q+m,cmp);
        rep(i,0,m-1)
        {
            int u=all(q[i].a),v=all(q[i].b);
            if(u!=v)
            {
                p[u]=v;
                ans[0]+=q[i].l;
                edge[q[i].a].pb({q[i].b,q[i].l}),edge[q[i].b].pb({q[i].a,q[i].l});
                cnt++;
                if(cnt==n-1)break;
            }
        }
        dfs(1);
        double y=1.0*(n-1)*n/2;
        printf("%.0f %.2f
",ans[0],ans[1]/y);
    }
    //system ("pause");
    return 0;
}

n(n≤100000)