网络流之费用流问题

费用流即最小费用最大流

先贴上粉书上的模板：

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(int u,int v,int c,int f,int w):
        from(u),to(v),cap(c),flow(f),cost(w)
        {}
};

int n,m;
vector<Edge> edges;
vector<int> G[maxn];
int inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];

void init(int n)
{
    //this->n=n;    粉书上原版有这一句，但明显会报错。感觉去掉也没什么影响...
    for (int i=0;i<n;i++)
        G[i].clear();
    edges.clear();
}

void AddEdge(int from,int to,int cap,int cost)
{
    edges.push_back(Edge(from,to,cap,0,cost));
    edges.push_back(Edge(to,from,0,0,-cost));
    m=edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}

bool BellmanFord(int s,int t,int &flow,long long &cost)        //用队列优化过，感觉和SPFA一样了
{
    for (int i=0;i<n;i++)   d[i]=INF;
    memset(inq,0,sizeof(inq));
    d[s]=0;     inq[s]=1;       p[s]=0;     a[s]=INF;

    queue<int> Q;
    Q.push(s);
    while (!Q.empty())
    {
        int u=Q.front();
        Q.pop();
        inq[u]=0;
        for (int i=0;i<G[u].size;i++)
        {
            Edge& e=edges[G[u][i]];
            if (e.cap>e.flow && d[e.to]>d[u]+e.cost())
            {
                d[e.to]=d[u]+e.cost;
                p[e.to]=G[u][i];
                a[e.to]=min(a[u],e.cap-e.flow);
                if (!inq[e.to])
                {
                    Q.push(e.to);
                    inq[e.to]=1;
                }
            }
        }
    }
    if (d[t]==INF)      return false;
    flow+=a[t];
    cost+=(long long)d[t] * (long long)a[t];
    for (int u=t;u!=s;u=edges[p[u]].from)
    {
        edges[p[u]].flow+=a[t];
        edges[p[u]^1].flow-=a[t];
    }
    return true;
}

int MCMF(int s,int t,long long& cost)        //s:起点；t:汇点；
{
    int flow=0;     cost=0;            //flow:求出来的最大流      cost:求出来的费用
    while (BellmanFord(s,t,flow,cost));
    return flow;
}

Exercise：POJ2516

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