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这是我在StackOverflow上的第一篇文章，所以请告诉我如果我做错了什么，而且英文不是我的母语，请原谅我，如果有任何格式化错误。从GoogleMapsAPI NuGet包中排列类型为“位置”的数组排列
我的问题是如何排列“位置”类型数组的项目，我需要获取用户给出的所有路标的排列，然后根据时间或距离计算出最佳路线。 （我不想使用正常的路线计算）
我已经搜索了算法，但是当我把类型“Location []”的数组放在函数的参数中时，我得到的错误是对象需要IEnumerable，我不知道如何转换，如果甚至可能的话，我从来没有使用过IEnumerable。
如果它的任何帮助，这是我的计算路线代码：
//Gets the waypoints from a listBox provided by the user, "mode" selects between best time and best distance
//backgroundworker so the UI dont freezes, and return the optimal waypoint order
public Location[] CalcularRota(Location[] waypoints, int mode, BackgroundWorker work, DoWorkEventArgs e)
{
//Declarations
string origem = "";
string destino = "";
Rota[] prop = new Rota[100]; //this index is the number of times the algorithm will be executed, more equals accuracy but much more time to complete
Rota bestDist = new Rota();
Rota bestTime = new Rota();
DirectionService serv = new DirectionService();
DirectionRequest reqs = new DirectionRequest();
DirectionResponse resp;
Random rnd = new Random();
Location[] rndWays;
int dist = 0;
int ti = 0;
bestDist.Distance = 1000000000; //put higher values for the first comparation to be true (end of code)
bestTime.Time = 1000000000;
if (waypoints != null)
{
reqs.Sensor = false;
reqs.Mode = TravelMode.driving;
for (int i = 0; i < prop.Length; i++) //initializes prop
prop[i] = new Rota();
for (int i = 0; i < prop.Length; i++)
{
rndWays = waypoints.OrderBy(x => rnd.Next()).ToArray(); //randomizes the order, I want to get all permutations and then test
//but I dont know how so I've been using randomized
dist = ti = 0;
origem = prop[0].ToString(); //save this particular waypoint's origin and destination
destino = prop[1].ToString();
reqs.Origin = origem;
reqs.Destination = destino;
if (waypoints.Length > 0)
reqs.Waypoints = rndWays;
resp = serv.GetResponse(reqs); //request the route with X order of waypoints to google
if (resp.Status == ServiceResponseStatus.Ok) //wait the response otherwise the program crashes
{
for (int j = 0; j < resp.Routes[0].Legs.Length; j++) //gets the distance and time of this particular order
{
ti += int.Parse(resp.Routes[0].Legs[j].Duration.Value);
dist += int.Parse(resp.Routes[0].Legs[j].Distance.Value);
}
}
prop[i].Origem = origem; //saves this waypoints order details for further comparison
prop[i].Destino = destino;
prop[i].Distance = dist;
prop[i].Time = ti;
prop[i].Order = rndWays;
work.ReportProgress(i); //report the progress
}
for (int i = 0; i < prop.Length; i++) //gets the best distance and time
{
if (bestDist.Distance > prop[i].Distance)
{
bestDist.Distance = prop[i].Distance;
bestDist.Time = prop[i].Time;
bestDist.Order = prop[i].Order;
bestDist.Origem = prop[i].Origem;
bestDist.Destino = prop[i].Destino;
}
if (bestTime.Time > prop[i].Time)
{
bestTime.Distance = prop[i].Distance;
bestTime.Time = prop[i].Time;
bestTime.Order = prop[i].Order;
bestTime.Origem = prop[i].Origem;
bestTime.Destino = prop[i].Destino;
}
}
if (bestDist.Order == bestTime.Order) //if the same waypoint order has the same time and distance
return bestDist.Order; // returns whatever bestDist.Order or bestTime.Order
else if (bestDist.Order != bestTime.Order) //if different returns corresponding to the mode selected
{
if (mode == 1) return bestDist.Order;
if (mode == 2) return bestTime.Order;
}
}
return null;
}
我想是到重排列给出的航点和测试每个排列，我一直在挣扎这一段时间，如果你们能以任何方式帮助我，那就太好了。
Ty。
编辑。
我在这里发现这个功能在计算器上：
public static bool NextPermutation<T>(T[] elements) where T : IComparable<T>
{
var count = elements.Length;
var done = true;
for (var i = count - 1; i > 0; i--)
{
var curr = elements[i];
// Check if the current element is less than the one before it
if (curr.CompareTo(elements[i - 1]) < 0)
{
continue;
}
// An element bigger than the one before it has been found,
// so this isn't the last lexicographic permutation.
done = false;
// Save the previous (bigger) element in a variable for more efficiency.
var prev = elements[i - 1];
// Have a variable to hold the index of the element to swap
// with the previous element (the to-swap element would be
// the smallest element that comes after the previous element
// and is bigger than the previous element), initializing it
// as the current index of the current item (curr).
var currIndex = i;
// Go through the array from the element after the current one to last
for (var j = i + 1; j < count; j++)
{
// Save into variable for more efficiency
var tmp = elements[j];
// Check if tmp suits the "next swap" conditions:
// Smallest, but bigger than the "prev" element
if (tmp.CompareTo(curr) < 0 && tmp.CompareTo(prev) > 0)
{
curr = tmp;
currIndex = j;
}
}
// Swap the "prev" with the new "curr" (the swap-with element)
elements[currIndex] = prev;
elements[i - 1] = curr;
// Reverse the order of the tail, in order to reset it's lexicographic order
for (var j = count - 1; j > i; j--, i++)
{
var tmp = elements[j];
elements[j] = elements[i];
elements[i] = tmp;
}
// Break since we have got the next permutation
// The reason to have all the logic inside the loop is
// to prevent the need of an extra variable indicating "i" when
// the next needed swap is found (moving "i" outside the loop is a
// bad practice, and isn't very readable, so I preferred not doing
// that as well).
break;
}
// Return whether this has been the last lexicographic permutation.
return done;
}
用法是：
NextPermutation(array);
这样做的，把我的阵列（rndWays）的过载，我得到了以下错误：
类型'Google.Maps.Location'不能用作泛型类型或方法'Form1.NextPermutation < T>（T []）'中的类型参数'T''。没有从'Google.Maps.Location'到'System.IComparable < Google.Maps.Location>'的隐式引用转换。
C＃中的数组应该实现IEnumerable。你能提供特定的代码来显示错误吗？ –
添加置换算法。 –