Hello codeforces! so as we know Floyd–Warshals algorithm finds the shortest path between all pairs of vertices and Bellman-Ford finds the shortest path from one note to all and if number of vertices will be_V and number of edges E then if we will run bellman ford from all vertices except one it will take o((v-1)*v*e) and floyd warshall takes o(v*v*v) that is mostly more than o((v-1)*v*e) am I right? So why do we need floyd warshalls algorithm?
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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 16.06.2024 06:11:28 (k3).
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Let, $$$n$$$ = no of nodes, $$$m$$$ = no of edges Finding all pair shortest path using belmenford takes $$$O(n^2m)$$$ but floyed warshal takes $$$O(n^3)$$$ time. floyed is very easy to write and code is short.
Floyd can be optimized to n^3/32
using bitset?
Actually you shouldn't mention about constant time complexity. That's useless.
What do you mean by $$$O(V^3)$$$ is mostly more than $$$O(V^2E)$$$?
What if $$$E\gg V$$$?
Yes that why I said mostly and I think that bellman ford takes o((v-1)*v*e)
So then in these cases where $$$E\gg V$$$, the Floyd-Warshall will be much faster, so that's why it's useful
It's useful when E is much bigger than V