hdu 5723 Abandoned country 最小生成树+子节点统计

Abandoned country

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3006    Accepted Submission(s): 346

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

 

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1e6+10;
int num[N],f[N];
vector<int> G[N];

struct edge{
   int u,v,cost,flag;
}e[N];

bool cmp(edge a,edge b)
{
   return a.cost<b.cost;
}

int findr(int u)
{
    if(f[u]!=u)
        f[u]=findr(f[u]);
    return f[u];
}

int dfs_clock;

void dfs(int u,int pre)
{
    int p=++dfs_clock;
    for(int i=0;i<G[u].size();i++)
    {
        int v=G[u][i];
        if(v==pre) continue;
        dfs(v,u);
    }
    num[u]=dfs_clock-p+1;
}

int main()
{
    int cas,n,m;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) G[i].clear();
        for(int i=1;i<=m;i++)
                {
                    scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].cost);
                     e[i].flag=0;
                }
        sort(e+1,e+m+1,cmp);
        for(int i=1;i<=n;i++) f[i]=i;
        ll cost=0;
        for(int i=1;i<=m;i++)
        {
            int u=findr(e[i].u),v=findr(e[i].v);
            if(u==v) continue;
            f[u]=v;
            e[i].flag=1;
            cost+=e[i].cost;
            G[e[i].u].push_back(e[i].v);
            G[e[i].v].push_back(e[i].u);
        }

        dfs_clock=0;
        dfs(1,-1);
        double q=0;
        for(int i=1;i<=m;i++)
           if(e[i].flag)
           {
               int u=e[i].u,v=e[i].v;
               ll k=min(num[u],num[v]);
               q+=k*(n-k)*e[i].cost;
           }

       q=2*q/n/(n-1);
       printf("%lld %.2f
",cost,q);
    }
    return 0;
}

　　分析:刚开始以为是道kruskal+统计子节点的水题，，，，后来写了下，，发现这道题目有个很迷的地方，就是连接起所有村庄的最小的总cost和最小的期望值，最小的cost当然跑kruskal，那会不会在同一种

cost的情况下出现不同期望值？那该怎么选择，，，于是瞬间懵逼，，就放弃了。。。

可见，，比赛时缺乏基本的随机应变和见招拆招的能力，，这要靠比赛来加强了。。。。比赛时胆子真的太小了，，，只有多多比赛慢慢克服了。。。

解决：其实因为题目说了所有边的权值均不同，，kruskal中对边sort后，形成最小生成树的边的选取方案是

唯一的，所以最小生成树唯一；