Nature Reserve(最小生成森林)

Nature Reserve

Kattis - naturereserve

In a Nature Reserve and Wildlife Park, there are $N$ environmental monitoring stations to monitor temperature, atmospheric pressure, humidity, fire, water quality, etc. Each station, labeled from $1$ to $N$, uses solar panels to supply energy for its operations. There is a communication network consisting of several two-way communication channels between pairs of stations. All stations are connected via this communication network.

To process data at each station, the Nature Reserve and Wildlife Park needs to install a Smart Data Analysis program (with the size of $L$ bytes) to all environmental monitoring stations. The program is initially installed directly to $S$ stations, then broadcast to and installed in all other stations via the communication network.

To save energy, all communication channels are initially in an idle state and need to be activated to send information. It takes ${E}_{ij}$ energy units to activate the communication channel between station $i$ and station $j$. Once a channel is activated, it takes one energy unit to transmit one byte via this channel.

Your task is to determine the minimum energy units required to send the Smart Data Analysis program to all stations from the initial $S$ stations.

Input
The input consists of several datasets. The first line of the input contains the number of datasets, which is a positive number and is not greater than $20$. The following lines describe the datasets.

Each dataset is described by the following lines:

The first line contains four positive integers: the number of environmental monitoring stations $N$, the number of two-way communication channels $M$, the size of the program $L$ (in bytes), and the number of initial stations $S$. The constraints are $1 \leq S \leq N \leq {10}^{4}$, $1 \leq M \leq {10}^{6}$, $M \leq N(N-1)/2$, and $1 \leq L \leq {10}^{6}$.

The second line contains $S$ positive integer representing the initial $S$ stations.

Each of the following $M$ lines contains three positive integers $i, j$ and ${E}_{ij}$ to denote that there is a two-way communicataion channel between station $i$ and station $j$, and it takes ${E}_{ij}$ energy units to activate this channel $({E}_{ij} \leq {10}^{6})$.

Output
For each data set, output the minimum energy units required to send the Smart Data Analysis program to all stations from the initial $S$ stations.

Sample Input 1
1
4 6 10 1
3
1 2 4
1 3 8
1 4 1
2 3 2
2 4 5
3 4 20

Sample Output 1
37

題意：n個點m條邊的無向圖，有s個初始點有文件，要把s個點的文件通過邊傳到其它所有點，每傳一個文件，耗費L,每條邊第一次走，要耗費這條邊的權值，問最少耗費是多少

• 構造s個最小生成樹，用並查集判斷，每個並查集的首領爲這s個點中的一個

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4+9;
long long  fa[maxn],flag[maxn];
struct node
{
    long long  s,e,v;
} edge[maxn*100];
long long root(long long a)
{
    if(fa[a]==a)
        return a;
    return fa[a] = root(fa[a]);
}
void join(long long a,long long b)
{
    long long ffa= root(a);
    long long ffb = root(b);
    if(ffa!=ffb)
    {
        if(flag[ffa])
            fa[ffb] = ffa;
        else
            fa[ffa] = ffb;
    }
}
bool cmp(node a,node b)
{
    return a.v<b.v;
}
int main()
{
    long long i,j,m,n,l,t,s,num,ans;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld %lld %lld %lld",&n,&m,&l,&s);
        for(i = 1; i<=n; i++)
        {
            fa[i] = i;
            flag[i] = 0;
        }
        for(i = 0; i<s; i++)
        {
            scanf("%lld",&num);
            flag[num] = 1;
        }
        for(i = 0; i<m; i++)
        {
            scanf("%lld %lld %lld",&edge[i].s,&edge[i].e,&edge[i].v);
        }

        sort(edge,edge+m,cmp);
        ans = 0;
        for(i = 0; i<m; i++)
        {
            long long sta = edge[i].s;
            long long endd = edge[i].e;
            long long val = edge[i].v;
            if(flag[root(sta)]&&flag[root(endd)]||root(sta)==root(endd))
                continue;
            else
            {
                join(sta,endd);
                ans += val+l;
            }
        }
        printf("%lld\n",ans);

    }
    return 0;
}