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In der Analyse Hopcroft und Karp Algorithmus haben wir gesehen, dass wir nach ⌈(√|V|) / 2⌉ Durchläufen maximal nochmal (√|V|) brauchen. Wir können das gleiche aussagen, falls der kürzeste gefundene Pfad nach dem 10. Durchlauf schon Länge (√|V|) hat. (r/f)
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