最大流 最小费

最大流

int d[N];
bool bfs()
{
    static queue q;
    memset(d, 0x3c, sizeof d);
    d[s] = 1;
    q.push(s);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        for(int e=fst[u]; e; e=nxt[e])
        {
            int v = to[e];
            if(!d[v] && c[e])
            {
                d[v] = d[u]+1;
                q.push(v);
            }
        }
    }
}

int dfs(int u, int flow)
{
    if(u==t) return flow;
    int rest  = flow, out;
    for(int e=fst[u];e&&rest;e=nxt[e])
    {
        int v = to[e];
        if(d[v]==d[u]+1 && c[e]) //still c[e
        {
            if(out = dfs(v, min(rest, c[e]))) 
                c[e]-=out, c[e^1]+=out, rest-=out;
            else d[v]=0;
        }
    }
    return flow-rest;
}

while(bfs()) ans+=dfs(s, INF);

aE(u,v,c); aE(v,u,0);

最小费用最大流

namespace mfmc
{
    int d[N];
    bool bfs()
    {
        static queue q; q.push(s);
        static bool inq[N]; memset(inq, 0, sizeof inq); inq[s]=1;
        memset(d, 0x3f, sizeof d); d[s]=0;
        memset(f, 0x3f, sizeof f);
        fa[t]=-1;
        while(!q.empty())
        {
            int u = q.front(); q.pop(); inq[u]=0;
            for(int e=fst[u];e;e=nxt[e])
            {
                int v = to[e];
                if(c[e]&&d[v]>d[u]+cost[e])
                {
                    d[v]=d[u]+cost[e];
                    if(!inq[v]) inq[v]=1, q.push(v);
                    f[v] = min(f[u], c[e]);
                    fa[v] = u;
                    ine[v] = e;
                }
            }
        }
        return ~fa[t];
    }
    int mc,mf;
    void run()
    {
        while(bfs())
        {
            mc+=d[t]*f[t]; mf+=f[t];
            for(int u=t; u^s; u=fa[u]) c[ine[e]]-=f[t], c[ine[u]^1]+=f[t];
        }
    }
}