MaxFlow

Constructor

mf_graph<Cap> graph(int n)

It creates a graph of n vertices and $0$ edges. Cap is the type of the capacity.

Constraints

$0 \leq n \leq 10^8$
Cap is int or ll

Complexity

$O(n)$

add_edge

int graph.add_edge(int from, int to, Cap cap);

It adds an edge oriented from the vertex from to the vertex to with the capacity cap and the flow amount $0$. It returns an integer $k$ such that this is the $k$-th edge that is added.

Constraints

$0 \leq \mathrm{from}, \mathrm{to} \lt n$
$0 \leq \mathrm{cap}$

Complexity

$O(1)$ amortized

flow

(1) Cap graph.flow(int s, int t);
(2) Cap graph.flow(int s, int t, Cap flow_limit);

It augments the flow from $s$ to $t$ as much as possible. It returns the amount of the flow augmented.
You may call it multiple times. See Appendix for further details.

Constraints

$s \neq t$
The answer should be in Cap.

Complexity

$O(\min(n^{\frac{2}{3}}m, m^{\frac{3}{2}}))$ (if all the capacities are $1$) or
$O(n^2 m)$ (general),

where $m$ is the number of added edges.

min_cut

vector<bool> graph.min_cut(int s)

It returns a vector of length $n$, such that the $i$-th element is true if and only if there is a directed path from $s$ to $i$ in the residual network. The returned vector corresponds to a $s-t$ minimum cut after calling flow(s, t) exactly once without flow_limit. See Appendix for further details.

Complexity

$O(n + m)$, where $m$ is the number of added edges.

get_edge / edges

struct mf_graph<Cap>::edge {
    int from, to;
    Cap cap, flow;
};

(1) mf_graph<Cap>::edge graph.get_edge(int i);
(2) vector<mf_graph<Cap>::edge> graph.edges();

It returns the current internal state of the edges.
The edges are ordered in the same order as added by add_edge.

Constraints

Complexity

(1): $O(1)$
(2): $O(m)$, where $m$ is the number of added edges.

change_edge

void graph.change_edge(int i, Cap new_cap, Cap new_flow);

It changes the capacity and the flow amount of the $i$-th edge to new_cap and new_flow, respectively. It doesn't change the capacity or the flow amount of other edges. See Appendix for further details.

Constraints

$0 \leq \mathrm{newflow} \leq \mathrm{newcap}$

Complexity

$O(1)$

Examples

AC code of https://atcoder.jp/contests/practice2/tasks/practice2_d

include atcoder/header import atcoder/maxflow let n, m = nextInt() var grid = newSeqWith(n, nextString()) # generate (s -> even grid -> odd grid -> t) graph # grid(i, j) correspond to vertex (i * m + j) var g = initMFGraph[int](n * m + 2) let s = n * m t = n * m + 1; # s -> even / odd -> t for i in 0..<n: for j in 0..<m: if grid[i][j] == '#': continue let v = i * m + j if (i + j) mod 2 == 0: g.add_edge(s, v, 1) else: g.add_edge(v, t, 1) # even -> odd for i in 0..<n: for j in 0..<m: if (i + j) mod 2 == 1 or grid[i][j] == '#': continue let v0 = i * m + j if i != 0 and grid[i - 1][j] == '.': let v1 = (i - 1) * m + j g.add_edge(v0, v1, 1) if j != 0 and grid[i][j - 1] == '.': let v1 = i * m + (j - 1) g.add_edge(v0, v1, 1) if i + 1 < n and grid[i + 1][j] == '.': let v1 = (i + 1) * m + j g.add_edge(v0, v1, 1) if j + 1 < m and grid[i][j + 1] == '.': let v1 = i * m + (j + 1) g.add_edge(v0, v1, 1) echo g.flow(s, t) let edges = g.edges() for e in edges: if e.src == s or e.dst == t or e.flow == 0: continue let i0 = e.src div m j0 = e.src mod m i1 = e.dst div m j1 = e.dst mod m if i0 == i1 + 1: grid[i1][j1] = 'v'; grid[i0][j0] = '^' elif j0 == j1 + 1: grid[i1][j1] = '>'; grid[i0][j0] = '<' elif i0 == i1 - 1: grid[i0][j0] = 'v'; grid[i1][j1] = '^' else: grid[i0][j0] = '>'; grid[i1][j1] = '<' echo grid.join("\n")