我实现Edmond Karp algorithm,但似乎不正确,我没有得到正确的流程,请考虑关注图表并从4到8流程:
算法运行如下:
首先发现4→1→8, 然后找到4→5→8 之后4→1→6→8
我认为第三条路径是错误的,因为使用这条路径我们不能使用6→8的流量(因为它使用了),实际上我们不能使用4→5→6→8的流量。
事实上,如果我们选择4→5→6→8,然后4→1→3→7→8然后4→1→3→7→8,我们可以获得更好的流量(40)。
我从wiki示例代码实现了算法。我认为我们不能使用任何有效的路径,事实上这种贪婪的选择是错误的。
我错了吗?
代码如下(在c#中,阈值为0,并且不影响算法):
public decimal EdmondKarps(decimal[][] capacities/*Capacity matrix*/,
List<int>[] neighbors/*Neighbour lists*/,
int s /*source*/,
int t/*sink*/,
decimal threshold,
out decimal[][] flowMatrix
/*flowMatrix (A matrix giving a legal flowMatrix with the maximum value)*/
)
{
THRESHOLD = threshold;
int n = capacities.Length;
decimal flow = 0m; // (Initial flowMatrix is zero)
flowMatrix = new decimal[n][]; //array(1..n, 1..n) (Residual capacity from u to v is capacities[u,v] - flowMatrix[u,v])
for (int i = 0; i < n; i++)
{
flowMatrix[i] = new decimal[n];
}
while (true)
{
var path = new int[n];
var pathCapacity = BreadthFirstSearch(capacities, neighbors, s, t, flowMatrix, out path);
if (pathCapacity <= threshold)
break;
flow += pathCapacity;
//(Backtrack search, and update flowMatrix)
var v = t;
while (v != s)
{
var u = path[v];
flowMatrix[u][v] = flowMatrix[u][v] + pathCapacity;
flowMatrix[v][u] = flowMatrix[v][u] - pathCapacity;
v = u;
}
}
return flow;
}
private decimal BreadthFirstSearch(decimal[][] capacities, List<int>[] neighbors, int s, int t, decimal[][] flowMatrix, out int[] path)
{
var n = capacities.Length;
path = Enumerable.Range(0, n).Select(x => -1).ToArray();//array(1..n)
path[s] = -2;
var pathFlow = new decimal[n];
pathFlow[s] = Decimal.MaxValue; // INFINT
var Q = new Queue<int>(); // Q is exactly Queue :)
Q.Enqueue(s);
while (Q.Count > 0)
{
var u = Q.Dequeue();
for (int i = 0; i < neighbors[u].Count; i++)
{
var v = neighbors[u][i];
//(If there is available capacity, and v is not seen before in search)
if (capacities[u][v] - flowMatrix[u][v] > THRESHOLD && path[v] == -1)
{
// save path:
path[v] = u;
pathFlow[v] = Math.Min(pathFlow[u], capacities[u][v] - flowMatrix[u][v]);
if (v != t)
Q.Enqueue(v);
else
return pathFlow[t];
}
}
}
return 0;
}
答案 0 :(得分:2)
选择路径的方法并不重要。
您必须以相反的顺序添加路径边缘和路径容量,并通过该值减少路径边缘的容量。
实际上这个解决方案有效:
while there is a path with positive capacity from source to sink{
find any path with positive capacity from source to sink, named P with capacity C.
add C to maximum_flow_value.
reduce C from capacity of edges of P.
add C to capacity of edges of reverse_P.
}
最后,最大流量的值是循环中C
s的总和。
如果要查看最大流量边缘的流量,可以将初始图形保留在某处,边缘e中的流量为original_capacity_e - current_capacity_e。