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Max Flow (Dinic's, HLPP)
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template <int MAXV, class T = int> struct Dinic { | |
const static bool SCALING = false; // non-scaling = V^2E, Scaling=VElog(U) with higher constant | |
int lim = 1; | |
const T INF = numeric_limits<T>::max(); | |
struct edge { | |
int to, rev; | |
T cap, flow; | |
}; | |
int s = MAXV - 2, t = MAXV - 1; | |
int level[MAXV], ptr[MAXV]; | |
vector<edge> adj[MAXV]; | |
void addEdge(int a, int b, T cap, bool isDirected = true) { | |
adj[a].push_back({b, adj[b].size(), cap, 0}); | |
adj[b].push_back({a, adj[a].size() - 1, isDirected ? 0 : cap, 0}); | |
} | |
bool bfs() { | |
queue<int> q({s}); | |
fill(all(level), -1); | |
level[s] = 0; | |
while (!q.empty() && level[t] == -1) { | |
int v = q.front(); | |
q.pop(); | |
for (auto e : adj[v]) { | |
if (level[e.to] == -1 && e.flow < e.cap && (!SCALING || e.cap - e.flow >= lim)) { | |
q.push(e.to); | |
level[e.to] = level[v] + 1; | |
} | |
} | |
} | |
return level[t] != -1; | |
} | |
T dfs(int v, T flow) { | |
if (v == t || !flow) | |
return flow; | |
for (; ptr[v] < adj[v].size(); ptr[v]++) { | |
edge &e = adj[v][ptr[v]]; | |
if (level[e.to] != level[v] + 1) | |
continue; | |
if (T pushed = dfs(e.to, min(flow, e.cap - e.flow))) { | |
e.flow += pushed; | |
adj[e.to][e.rev].flow -= pushed; | |
return pushed; | |
} | |
} | |
return 0; | |
} | |
long long calc() { | |
long long flow = 0; | |
for (lim = SCALING ? (1 << 30) : 1; lim > 0; lim >>= 1) { | |
while (bfs()) { | |
fill(all(ptr), 0); | |
while (T pushed = dfs(s, INF)) | |
flow += pushed; | |
} | |
} | |
return flow; | |
} | |
}; |
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template <int MAXN, class T = int> struct HLPP { | |
const T INF = numeric_limits<T>::max(); | |
struct edge { | |
int to, rev; | |
T f; | |
}; | |
int s = MAXN - 1, t = MAXN - 2; | |
vector<edge> adj[MAXN]; | |
vector<int> lst[MAXN], gap[MAXN]; | |
T excess[MAXN]; | |
int highest, height[MAXN], cnt[MAXN], work; | |
void addEdge(int from, int to, int f, bool isDirected = true) { | |
adj[from].push_back({to, adj[to].size(), f}); | |
adj[to].push_back({from, adj[from].size() - 1, isDirected ? 0 : f}); | |
} | |
void updHeight(int v, int nh) { | |
work++; | |
if (height[v] != MAXN) | |
cnt[height[v]]--; | |
height[v] = nh; | |
if (nh == MAXN) | |
return; | |
cnt[nh]++, highest = nh; | |
gap[nh].push_back(v); | |
if (excess[v] > 0) | |
lst[nh].push_back(v); | |
} | |
void globalRelabel() { | |
work = 0; | |
fill(all(height), MAXN); | |
fill(all(cnt), 0); | |
for (int i = 0; i < highest; i++) | |
lst[i].clear(), gap[i].clear(); | |
height[t] = 0; | |
queue<int> q({t}); | |
while (!q.empty()) { | |
int v = q.front(); | |
q.pop(); | |
for (auto &e : adj[v]) | |
if (height[e.to] == MAXN && adj[e.to][e.rev].f > 0) | |
q.push(e.to), updHeight(e.to, height[v] + 1); | |
highest = height[v]; | |
} | |
} | |
void push(int v, edge &e) { | |
if (excess[e.to] == 0) | |
lst[height[e.to]].push_back(e.to); | |
T df = min(excess[v], e.f); | |
e.f -= df, adj[e.to][e.rev].f += df; | |
excess[v] -= df, excess[e.to] += df; | |
} | |
void discharge(int v) { | |
int nh = MAXN; | |
for (auto &e : adj[v]) { | |
if (e.f > 0) { | |
if (height[v] == height[e.to] + 1) { | |
push(v, e); | |
if (excess[v] <= 0) | |
return; | |
} else | |
nh = min(nh, height[e.to] + 1); | |
} | |
} | |
if (cnt[height[v]] > 1) | |
updHeight(v, nh); | |
else { | |
for (int i = height[v]; i <= highest; i++) { | |
for (auto j : gap[i]) | |
updHeight(j, MAXN); | |
gap[i].clear(); | |
} | |
} | |
} | |
T calc(int heur_n = MAXN) { | |
fill(all(excess), 0); | |
excess[s] = INF, excess[t] = -INF; | |
globalRelabel(); | |
for (auto &e : adj[s]) | |
push(s, e); | |
for (; highest >= 0; highest--) { | |
while (!lst[highest].empty()) { | |
int v = lst[highest].back(); | |
lst[highest].pop_back(); | |
discharge(v); | |
if (work > 4 * heur_n) | |
globalRelabel(); | |
} | |
} | |
return excess[t] + INF; | |
} | |
}; |
One limitation of the HLPP implementation is that you can't recover the weights for the full flow - use Dinic's for this.
actually what means scaling here?
This is the only hlpp solution I can find on the web.
hlpp.cpp is buggy, as mentioned in a comment on Codeforces.
On the following max flow test case (8 nodes, 8 directed edges, source = 1, sink = 7)
8 8 1 7
1 7 1
1 2 2
2 3 1
3 4 1
4 5 1
5 6 1
6 7 1
8 3 1
hlpp.cpp outputs maxflow = 1, but the correct answer is 2.
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Both of these implementations are very fast, although the HLPP one is faster. On all of these problems, this HLPP implementation is the fastest submission/tied for the fastest.
Dinic's:
SPOJ FASTFLOW (ran without scaling): 0.04
SPOJ FASTFLOW (ran with scaling): 0.27
VNSPOJ FASTFLOW (ran with scaling): 0.00 (1st!)
VNSPOJ FASTFLOW (ran without scaling): TLE
Weird Chinese Flow Problem that's trying to force you to use HLPP (https://loj.ac/problem/127): 88 points with scaling, 44 without
HLPP:
SPOJ FASTFLOW: 0.00 (1st)
VNSPOJ FASTFLOW: 0.00 (1st)
LOJ: 100 pts, 1087 ms (1st)