最小費用最大流(白書+kuangbin)
阿新 • • 發佈:2018-10-31
講解:
附上白書程式碼:
#include<bits/stdc++.h> using namespace std; const int MAX_V=100; const int INF=0x3f3f3f3f; struct edge{ int to,cap,cost,rev; edge(int _to,int _cap,int _cost,int _rev):to(_to),cap(_cap),cost(_cost),rev(_rev){} }; int V; vector<edge>G[MAX_V]; int dist[MAX_V]; int prevv[MAX_V],preve[MAX_V]; void add_edge(int from,int to,int cap,int cost) { G[from].push_back(edge(to,cap,cost,G[to].size())); G[to].push_back(edge(from,0,-cost,G[from].size()-1)); } int min_cost_flow(int s,int t,int f) { int res=0; while(f>0){ fill(dist,dist+V,INF); dist[s]=0; bool update=true; while(update){ update=false; for(int v=0;v<V;v++){ if(dist[v]==INF){ continue; } for(int i=0;i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&dist[e.to]>dist[v]+e.cost){ dist[e.to]=dist[v]+e.cost; prevv[e.to]=v; preve[e.to]=i; update=true; } } } } if(dist[t]==INF){ return -1; } int d=f; for(int v=t;v!=s;v=prevv[v]){ d=min(d,G[prevv[v]][preve[v]].cap); } f-=d; res+=d*dist[t]; for(int v=t;v!=s;v=prevv[v]){ edge &e=G[prevv[v]][preve[v]]; e.cap-=d; G[v][e.rev].cap+=d; } } return res; } int main() { int m,s,t,f; scanf("%d%d%d%d%d",&V,&m,&s,&t,&f); int u,v,w,cost; for(int i=0;i<m;i++){ scanf("%d%d%d%d",&u,&v,&w,&cost); add_edge(u,v,w,cost); } printf("%d\n",min_cost_flow(s,t,f)); return 0; }
附上白書Dijkstra程式碼:
#include<bits/stdc++.h> using namespace std; const int MAX_V=100; const int INF=0x3f3f3f3f; typedef pair<int,int>P; struct edge{ int to,cap,cost,rev; edge(int _to,int _cap,int _cost,int _rev):to(_to),cap(_cap),cost(_cost),rev(_rev){} }; int V; vector<edge>G[MAX_V]; int h[MAX_V]; int dist[MAX_V]; int prevv[MAX_V],preve[MAX_V]; void add_edge(int from,int to,int cap,int cost) { G[from].push_back(edge(to,cap,cost,G[to].size())); G[to].push_back(edge(from,0,-cost,G[from].size()-1)); } int min_cost_flow(int s,int t,int f) { int res=0; fill(h,h+V,0); while(f>0){ priority_queue<P,vector<P>,greater<P> >que; fill(dist,dist+V,INF); dist[s]=0; que.push(P(0,s)); while(!que.empty()){ P p=que.top();que.pop(); int v=p.second; if(dist[v]<p.first){ continue; } for(int i=0;i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){ dist[e.to]=dist[v]+e.cost+h[v]-h[e.to]; prevv[e.to]=v; preve[e.to]=i; que.push(P(dist[e.to],e.to)); } } } if(dist[t]==INF){ return -1; } for(int v=0;v<V;v++){ h[v]+=dist[v]; } int d=f; for(int v=t;v!=s;v=prevv[v]){ d=min(d,G[prevv[v]][preve[v]].cap); } f-=d; res+=d*h[t]; for(int v=t;v!=s;v=prevv[v]){ edge &e=G[prevv[v]][preve[v]]; e.cap-=d; G[v][e.rev].cap+=d; } } return res; } int main() { int m,s,t,f; scanf("%d%d%d%d%d",&V,&m,&s,&t,&f); int u,v,w,cost; for(int i=0;i<m;i++){ scanf("%d%d%d%d",&u,&v,&w,&cost); add_edge(u,v,w,cost); } printf("%d\n",min_cost_flow(s,t,f)); return 0; }
附上Kuangbin程式碼:
//spfa版費用流 //最小費用最大流,求最大流費用只需要取相反數,結果取相反數即可 //點的總數為N,點的編號為0~N-1 #include<bits/stdc++.h> using namespace std; const int MAXN=10000; const int MAXM=100000; const int INF=0x3f3f3f3f; struct Edge{ int to,next,cap,flow,cost; }edge[MAXM]; int head[MAXN],tol; int pre[MAXN],dis[MAXN]; bool vis[MAXN]; int N; void init(int n) { N=n; tol=0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int cap,int cost) { edge[tol].to=v; edge[tol].cap=cap; edge[tol].cost=cost; edge[tol].flow=0; edge[tol].next=head[u]; head[u]=tol++; edge[tol].to=u; edge[tol].cap=0; edge[tol].cost=-cost; edge[tol].flow=0; edge[tol].next=head[v]; head[v]=tol++; } bool spfa(int s,int t) { queue<int>q; for(int i=0;i<N;i++){ dist[i]=INF; vis[i]=false; pre[i]=-1; } dist[s]=0; vis[s]=true; q.push(s); while(!q.empty()){ int u=q.front(); q.pop(); vis[u]=false; for(int i=head[u];i!=-1;i=edge[i].next){ int v=edge[i].to; if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost){ dis[v]=dis[u]+edge[i].cost; pre[v]=i; if(!vis[v]){ vis[v]=true; q.push(v); } } } } if(pre[t]==-1){ return false; }else{ return true; } } int minCostMaxflow(int s,int t,int &cost) { int flow=0; cost=0; while(spfa(s,t)){ int Min=INF; for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){ if(Min>edge[i].cap-edge[i].flow){ Min=edge[i].cap-edge[i].flow; } } for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){ edge[i].flow+=Min; edge[i^1].flow-=Min; cost+=edge[i].cost*Min; } flow+=Min; } return flow; }