MaxFlow
It solves maximum flow problem.
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
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
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
Examples
#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;
}