March 7, 2011

[基础算法] 一个Ford-Fulkerson算法的例子

Example:
3-7-2011 10-07-02 AM 

Initialized:
2

Augmenting path:s→a→b→t,excess capacity:min(4, 3, 5) = 3
f(G) = 3:
3

Augmenting path:s→a→t,excess capacity:min(1, 5) = 1
f(G) = 4:
4

Augmenting path:s→c→d→t,excess capacity:min(6, 4, 4) = 4
f(G) = 8:
5 

Final result:
6

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