运筹学最大最小,运筹学最大流最小费用问题谁会( 四 ) _生活百科

w "(u, v) = 0.
If (u, v) rather than by the side of flow path, it must have:
L (v) ≤ L (u) + w (u, v),
Into the (*)-type, there w (u, v) ≥ 0.
Shows that the first amendment to w (e), against either side, there are w (e) ≥ 0, and a stream side (by side chain flow), there will be w (e) = 0. Calculated after each iteration, if f (u, v)> 0, by the need to establish the network flow (v, u) edge, edge weights w '(v, u) =- w (u, v) = 0, that is, the right will not be a negative side.
In addition, the calculation of each iteration with (*) fixes all the w (e), it is not difficult to prove that to each path x to y, its all the same to increase the path length L (x)-L (y). Therefore, x and y will not be the shortest path to w (e) the amendment changes.
】【Calculation steps
1. On the network G = [V, E, C, W], given the initial value is zero flow streams.
2. Be accompanied by this stream flow network G '= [V', E ', W'].
G 'the vertex-G: V' = V.
If G in f (u, v) If G in f (u, v)> 0, then G 'in the building edge (v, u), w' (v, u) =- w (u, v).
3. If G 'does not exist the path x to y, then G is the minimum cost flow of maximum flow,
To stop calculation; otherwise labeling method used to find x to y of the shortest path P.
4. According to P, the increased flow in G: each edge of P of (u, v), if G exists (u, v), then (u, v) by flow; if G exists (v, u), while (v, u) by flow. Increase (decrease) in flow should be on either side to ensure that there is c (e) ≥ f (e) ≥ 0.
5. According to the calculation of the shortest path at the time of peak value of the label L (v), press the G-type modification of all the edge weights w (e):
L (u)-L (v) + w (e) → w (e).
6. The new stream as the initial flow to 2.
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希望能满足您的要求----
运筹学。最小费用最大流中最后一步的总费用b怎么求?我已经求出了最小费用最大流f，但是b(f)怎么求。

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b(f) = sum(bij * fij), 即你求出的最大流各个边上流量fij 乘以单位流量费用 bij 求和。
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