O(n²+m)

int g[N][N], dist[N];//memset(g, 0x3f, sizeof g);
bool s[N];

int prim()//连通时返回最小生成树边权和，否则返回INF(0x3f3f3f3f)
{
    memset(dist, 0x3f, sizeof(dist));
    int res = 0;
    for (int i = 0; i < n; i++)
    {
        int t = -1;
        for (int j = 1; j <= n; j++)
        {
            if (!s[j] && (t == -1 || dist[t] > dist[j]))
            {
                t = j;
            }
        }
        if (i && dist[t] == INF)
        {
            return INF;
        }
        if (i)
        {
            res += dist[t];
        }
        for (int j = 1; j <= n; j++)
        {
            dist[j] = min(dist[j], g[t][j]);
        }
        s[t] = 1;
    }
    return res;
}