Results 1 
4 of
4
A Fast and Simple Algorithm for the Maximum Flow Problem
 OPERATIONS RESEARCH
, 1989
"... We present a simple sequential algorithm for the maximum flow problem on a network with n nodes, m arcs, and integer arc capacities bounded by U. Under the practical assumption that U is polynomially bounded in n, our algorithm runs in time O(nm + n 2 log n). This result improves the previous best b ..."
Abstract

Cited by 37 (6 self)
 Add to MetaCart
We present a simple sequential algorithm for the maximum flow problem on a network with n nodes, m arcs, and integer arc capacities bounded by U. Under the practical assumption that U is polynomially bounded in n, our algorithm runs in time O(nm + n 2 log n). This result improves the previous best bound of O(nm log(n 2 /m)), obtained by Goldberg and Taran, by a factor of log n for networks that are both nonsparse and nondense without using any complex data structures. We also describe a parallel implementation of the algorithm that runs in O(n'log U log p) time in the PRAM model with EREW and uses only p processors where p = [m/n
New DistanceDirected Algorithms for Maximum Flow and Parametric Maximum Flow Problems
, 1987
"... ..."
Max flows in O(nm) time, or better
, 2012
"... In this paper, we present improved polynomial time algorithms for the max flow problem defined on a network with n nodes and m arcs. We show how to solve the max flow problem in O(nm) time, improving upon the best previous algorithm due to King, Rao, and Tarjan, who solved the max flow problem in O( ..."
Abstract

Cited by 5 (0 self)
 Add to MetaCart
In this paper, we present improved polynomial time algorithms for the max flow problem defined on a network with n nodes and m arcs. We show how to solve the max flow problem in O(nm) time, improving upon the best previous algorithm due to King, Rao, and Tarjan, who solved the max flow problem in O(nm log m/(n log n) n) time. In the case that m = O(n), we improve the running time to O(n 2 / log n). We further improve the running time in the case that U ∗ = Umax/Umin is not too large, where Umax denotes the largest finite capacity and Umin denotes the smallest nonzero capacity. If log(U ∗ ) = O(n 1/3 log −3 n), we show how to solve the max flow problem in O(nm / log n) steps. In the case that log(U ∗ ) = O(log k n) for some fixed positive integer k, we show how to solve the max flow problem in Õ(n8/3) time. This latter algorithm relies on a subroutine for fast matrix multiplication. 1