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

(1): $0 \leq i \lt m$

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
#include <atcoder/maxflow>
#include <iostream>
using namespace std;
using namespace atcoder;
int main() {
    int n, m;
    cin >> n >> m;
    vector<string> grid(n);
    for (int i = 0; i < n; i++) {
        cin >> grid[i];
    }
    /**
     * generate (s -> even grid -> odd grid -> t) graph
     * grid(i, j) correspond to vertex (i * m + j)
     **/
    mf_graph<int> g(n * m + 2);
    int s = n * m, t = n * m + 1;
    // s -> even / odd -> t
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if (grid[i][j] == '#') continue;
            int v = i * m + j;
            if ((i + j) % 2 == 0) {
                g.add_edge(s, v, 1);
            } else {
                g.add_edge(v, t, 1);
            }
        }
    }
    // even -> odd
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if ((i + j) % 2 || grid[i][j] == '#') continue;
            int v0 = i * m + j;
            if (i && grid[i - 1][j] == '.') {
                int v1 = (i - 1) * m + j;
                g.add_edge(v0, v1, 1);
            }
            if (j && grid[i][j - 1] == '.') {
                int v1 = i * m + (j - 1);
                g.add_edge(v0, v1, 1);
            }
            if (i + 1 < n && grid[i + 1][j] == '.') {
                int v1 = (i + 1) * m + j;
                g.add_edge(v0, v1, 1);
            }
            if (j + 1 < m && grid[i][j + 1] == '.') {
                int v1 = i * m + (j + 1);
                g.add_edge(v0, v1, 1);
            }
        }
    }
    cout << g.flow(s, t) << endl;
    auto edges = g.edges();
    for (auto e : edges) {
        if (e.from == s || e.to == t || e.flow == 0) continue;
        int i0 = e.from / m, j0 = e.from % m;
        int i1 = e.to / m, j1 = e.to % m;
        if (i0 == i1 + 1) {
            grid[i1][j1] = 'v';
            grid[i0][j0] = '^';
        } else if (j0 == j1 + 1) {
            grid[i1][j1] = '>'; grid[i0][j0] = '<';
        } else if (i0 == i1 - 1) {
            grid[i0][j0] = 'v';
            grid[i1][j1] = '^';
        } else {
            grid[i0][j0] = '>'; grid[i1][j1] = '<';
        }
    }
    for (int i = 0; i < n; i++) {
        cout << grid[i] << endl;
    }
    return 0;
}
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX