题目链接：传送门

Description

n个和尚住的地方，m个其他地方。每个地方挖一口井需要花费q[i]的钱，两个地方连通需要花费对应的边权值。求所有人都能喝到水的最小花费。

Solution

新建一个节点，连接所有点，边权为挖井的花费，然后直接上斯坦纳树即可

Code

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#include <bits/stdc++.h>

using namespace std;

const int Maxn = 1000 + 10, Maxm = 10000 + 10, inf = 100000000;

int N, M, K;
int e, Begin[Maxm], To[Maxm * 2], Next[Maxm * 2], W[Maxm * 2];
int Dis[Maxn][(1 << 11)], Vis[Maxm], Dp[(1 << 11)];

inline void Init ()
{
    e = 0;
    memset(Begin, 0, sizeof Begin);
    memset(Vis, 0, sizeof Vis);
    for (int i = 1; i <= N + M; ++i)
        for (int j = 0; j < (1 << 10); ++j)
            Dis[i][j] = inf;
}

inline void add_edge (int x, int y, int z)
{
    To[++e] = y;
    Next[e] = Begin[x];
    Begin[x] = e;
    W[e] = z;
}

queue <int> Q;
inline void SPFA (int state)
{
    while (!Q.empty())
    {
        int x = Q.front(); Q.pop();
        Vis[x] = 0;
        for (int i = Begin[x]; i; i = Next[i])
        {
            int y = To[i];
            if (Dis[y][state] > Dis[x][state] + W[i])
            {
                Dis[y][state] = Dis[x][state] + W[i];
                if (!Vis[y])
                {
                    Vis[y] = 1;
                    Q.push(y);
                }
            }
        }
    }
}

int main()
{
#ifdef hk_cnyali
    freopen("A.in", "r", stdin);
    freopen("A.out", "w", stdout);
#endif
    while (~scanf("%d%d%d", &N, &M, &K))
    {
        Init();
        for (int i = 1; i <= N + M; ++i)
        {
            int x;
            scanf("%d", &x);
            if (i <= N)
            {
                Dis[i][(1 << (i - 1))] = 0;
                Dis[i][(1 << (i - 1)) | (1 << N)] = x;
            }
            else Dis[i][1 << N] = x;
        }
        while (K--)
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            add_edge(x, y, z);
            add_edge(y, x, z);
        }


        int tmp = N + 1;
        for (int state = 1; state < (1 << tmp); ++state)
        {
            for (int i = 1; i <= N + M; ++i)
            {
                for (int s1 = state & (state - 1); s1; s1 = state & (s1 - 1))
                    Dis[i][state] = min(Dis[i][state], Dis[i][(state - s1)] + Dis[i][s1]);
                if (Dis[i][state] != inf) Q.push(i), Vis[i] = 1;
            }
            SPFA(state);
        }

        for (int state = 1; state < (1 << tmp); ++state)
        {
            Dp[state] = inf;
            for (int i = 1; i <= N + M; ++i)
                Dp[state] = min(Dp[state], Dis[i][state]);//cout<<Dis[i][state]<<" ";
//          cout<<endl;
        }

        for (int state = 1; state < (1 << tmp); ++state)
        {
            if (!(state & (1 << N))) continue;
            for (int s1 = state & (state - 1); s1; s1 = state & (s1 - 1))
            {
                if (!(s1 & (1 << N))) continue;
                Dp[state] = min(Dp[state], Dp[s1] + Dp[(state - s1) | (1 << N)]);
            }
        }
        cout<<Dp[(1 << tmp) - 1]<<endl;
    }
    return 0;
}