下列 Dijkstra 算法，假设图 graph 中顶点数 $v$、边数 $e$，则程序的时间复杂度为（ ）。

typedef struct Edge {
    int in, out;    // 从下标in顶点到下标out顶点的边
    int len;        // 边长度
    struct Edge * next;
} Edge;
// v：顶点个数，graph：出边邻接表，start：起点下标，dis：输出每个顶点的最短距离
void dijkstra(int v, Edge * graph[], int start, int * dis) {
    const int MAX_DIS = 0x7fffff;
    for (int i = 0; i < v; i++)
        dis[i] = MAX_DIS;
    dis[start] = 0;
    int * visited = new int[v];
    for (int i = 0; i < v; i++)
        visited[i] = 0;
    visited[start] = 1;
    for (int t = 0; ; t++) {
        int min = MAX_DIS, minv = -1;
        for (int i = 0; i < v; i++) {
            if (visited[i] == 0 && min > dis[i]) {
                min = dis[i];
                minv = i;
            }
        }
        if (minv < 0)
            break;
        visited[minv] = 1;
        for (Edge * e = graph[minv]; e != NULL; e = e->next)
            if (dis[e->out] > e->len)
                dis[e->out] = e->len;
    }
    delete[] visited;
}

{{ select(1) }}

• $O(v^2)$
• $O(v \log v + e)$
• $O((v + e) \log v)$
• $O(v + e)$