多校HDU5723 最小生成树＋dfs回溯

Abandoned country

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 877    Accepted Submission(s): 236

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

题目大意：有ｎ（n<100000）个地点面m(m<1000000)条路，修每条路都有花费ｗ，求最小花费使每个地点能够互相到达并求出人一两点的花费期望；

因为ｎ很大所以用kruskal求最小生成树求出最小花费，然后dfs搜索回溯的办法找到所有情况每条路用过的次数并求出总花费，用总花费除以所有可能发生的次数（n*(n-1)/2）就是我们要求的期望。

其中用到vector容器进行dfs;

以下是代码：

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;
#define LL long  long
#define N 100010
#define M 1000010
vector<pair<int,int> >  v[N];//定义一个pair型的主要是因为要在v[].second中储存路径权值
struct node
{
    int a,b,w;
} edge[M];//储存每条边
int father[N],vis[N];
int n,m;
long long  ans;//记录总权值；
bool cmp(node x,node y){
    return x.w<y.w;
}
int finds(int x){
    return father[x]==x?x:father[x]=finds(father[x]);
}
long long dfs(int x){//dfs递归搜索
    vis[x]=1;　　　　　　　　　　　　　　　　　　　//标记顶点避免死循环
    int  i;
    long long nb=0,cb=0;
    int h=v[x].size();
    for(i=0; i<h; i++)　　　　　　　　　　　　　　
    {
        int b=v[x][i].first;　　　　　　　　　　
        if(!vis[b])
        {
            nb=dfs(b);
            cb+=nb;
            ans+=nb*(n-nb)*v[x][i].second;
        }
    }
    return 1+cb;
}
int main()
{
    int t,i;
    long long sum;
    cin>>t;
    while(t--)
    {
        for(i=0; i<=N; i++)
            v[i].clear();
        memset(vis,0,sizeof(vis));
        ans=0;
        sum=0;
        int flag=0;
        scanf("%d%d",&n,&m);
        if(n==0||m==0)
        {
            printf("0 0.00\n");
            continue;
        }
        for(i=1; i<=n; i++)
        {
            father[i]=i;
        }
        for(i=0; i<m; i++)
        {
            scanf("%d%d%d",&edge[i].a,&edge[i].b,&edge[i].w);
        }
        sort(edge,edge+m,cmp);
        for(i=0; i<m; i++)
        {
            int fx=finds(edge[i].a);
            int fy=finds(edge[i].b);
            if(fx!=fy)
            {
                flag++;
                father[fx]=fy;
                sum+=edge[i].w;
                v[edge[i].a].push_back(make_pair(edge[i].b,edge[i].w));
                v[edge[i].b].push_back(make_pair(edge[i].a,edge[i].w));
            }
            if(flag==n-1)
                break;
        }
        for(i=1; i<=n; i++)
        {
            if(v[i].size()==1)
                break;
        }
        long long  ko=dfs(i);
        double y=1.0*n*(n-1)/2;
        printf("%I64d %.2lf\n",sum,(double)ans/y);
    }
    return 0;
}