最小费用最大流模版(训练指南)
struct Edge { int from,to,cap,flow,cost; }; struct MCMF { int n,m,s,t; 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; REP(i,1,n) 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 bellman_ford(int s,int t,int &flow,int &cost) { REP(i,1,n) d[i]=INF; MS0(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]]; inq[u]=0; 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 0; flow+=a[t]; cost+=d[t]*a[t]; int u=t; while(u!=s){ edges[p[u]].flow+=a[t]; edges[p[u]^1].flow-=a[t]; u=edges[p[u]].from; } return 1; } int Mincost(int s,int t) { int flow=0,cost=0; while(bellman_ford(s,t,flow,cost)); return cost; } };MCMF F;
没有AC不了的题,只有不努力的ACMER!