Newtonsoft JSON Deserialize

You can implement a class that holds the fields you have in your JSON

class MyData
{
    public string t;
    public bool a;
    public object[] data;
    public string[][] type;
}

and then use the generic version of DeserializeObject:

MyData tmp = JsonConvert.DeserializeObject<MyData>(json);
foreach (string typeStr in tmp.type[0])
{
    // Do something with typeStr
}

Documentation: Serializing and Deserializing JSON

A much easier solution: Using a dynamic type

As of Json.NET 4.0 Release 1, there is native dynamic support. You don't need to declare a class, just use dynamic :

dynamic jsonDe = JsonConvert.DeserializeObject(json);

All the fields will be available:

foreach (string typeStr in jsonDe.Type[0])
{
    // Do something with typeStr
} 

string t = jsonDe.t;
bool a = jsonDe.a;
object[] data = jsonDe.data;
string[][] type = jsonDe.Type;

With dynamic you don't need to create a specific class to hold your data.