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Min Cost Max Flow - Dijkstra.cpp
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Min Cost Max Flow - Dijkstra.cpp
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//Works for negative costs, but does not work for negative cycles
//Complexity: O(min(E^2 *V log V, E logV * flow))
struct edge
{
int to, flow, cap, cost, rev;
};
struct MinCostMaxFlow
{
int nodes;
vector<int> prio, curflow, prevedge, prevnode, q, pot;
vector<bool> inqueue;
vector<vector<edge> > graph;
MinCostMaxFlow() {}
MinCostMaxFlow(int n): nodes(n), prio(n, 0), curflow(n, 0),
prevedge(n, 0), prevnode(n, 0), q(n, 0), pot(n, 0), inqueue(n, 0), graph(n) {}
void addEdge(int source, int to, int capacity, int cost)
{
edge a = {to, 0, capacity, cost, (int)graph[to].size()};
edge b = {source, 0, 0, -cost, (int)graph[source].size()};
graph[source].push_back(a);
graph[to].push_back(b);
}
void bellman_ford(int source, vector<int> &dist)
{
fill(dist.begin(), dist.end(), INT_MAX);
dist[source] = 0;
int qt=0;
q[qt++] = source;
for(int qh=0;(qh-qt)%nodes!=0;qh++)
{
int u = q[qh%nodes];
inqueue[u] = false;
for(auto &e : graph[u])
{
if(e.flow >= e.cap)
continue;
int v = e.to;
int newDist = dist[u] + e.cost;
if(dist[v] > newDist)
{
dist[v] = newDist;
if(!inqueue[v])
{
inqueue[v] = true;
q[qt++ % nodes] = v;
}
}
}
}
}
pair<int, int> minCostFlow(int source, int dest, int maxflow)
{
bellman_ford(source, pot);
int flow = 0;
int flow_cost = 0;
while(flow < maxflow)
{
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
q.push({0, source});
fill(prio.begin(), prio.end(), INT_MAX);
prio[source] = 0;
curflow[source] = INT_MAX;
while(!q.empty())
{
int d = q.top().first;
int u = q.top().second;
q.pop();
if(d != prio[u])
continue;
for(int i=0;i<graph[u].size();i++)
{
edge &e=graph[u][i];
int v = e.to;
if(e.flow >= e.cap)
continue;
int newPrio = prio[u] + e.cost + pot[u] - pot[v];
if(prio[v] > newPrio)
{
prio[v] = newPrio;
q.push({newPrio, v});
prevnode[v] = u;
prevedge[v] = i;
curflow[v] = min(curflow[u], e.cap - e.flow);
}
}
}
if(prio[dest] == INT_MAX)
break;
for(int i=0;i<nodes;i++)
pot[i]+=prio[i];
int df = min(curflow[dest], maxflow - flow);
flow += df;
for(int v=dest;v!=source;v=prevnode[v])
{
edge &e = graph[prevnode[v]][prevedge[v]];
e.flow += df;
graph[v][e.rev].flow -= df;
flow_cost += df * e.cost;
}
}
return {flow, flow_cost};
}
};
//Problem 1: https://www.spoj.com/problems/GREED/
//Solution 1: http://p.ip.fi/ODRk
//Problem 2 (Double Cost): https://codeforces.com/contest/277/problem/E
//Solution 2: https://codeforces.com/contest/277/submission/43180845