uestc WHITE ALBUM

思路：

只要连接到k个房间其中一个就可以，所以可以把k个房间并起来。

然后就是克鲁斯卡尔。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<vector>
#define INF 1e7
#define MAXN 100010
#define maxn 1000010
#define Mod 1000007
#define N 1010
using namespace std;
typedef long long LL;

struct node{
    int a, b, c;
    bool operator < (const node a) const{
        return c < a.c;
    }
}s[N*500];

int fa[N];
int n, m, k;

int findset(int x)
{
    if (fa[x] == x) return x;
    return fa[x] = findset(fa[x]);
}

void merg(int a, int b)
{
    int x = findset(a);
    int y = findset(b);
    if (x != y) {
        fa[x] = y;
    }
}

void init()
{
    for (int i = 0; i <= n; ++i)
        fa[i] = i;
}

int main()
{
    while (cin >> n >> m >> k) {
        init();
        int tmp, x;
        cin >> tmp;
        for (int i = 1; i < k; ++i) {
            cin >> x;
            fa[x] = tmp;
        }
        int a, b, c;
        for (int i = 0; i < m; ++i) {
            cin >> s[i].a >> s[i].b >> s[i].c;
        }
        sort(s,s+m);
        int res = 0;
        for (int i = 0; i < m; ++i) {
            int x = findset(s[i].a);
            int y = findset(s[i].b);
            if (x != y) {
                fa[x] = y;
                res += s[i].c;
            }
        }
        cout << res << endl;
    }
    return 0;
}