题解:先在原网络上跑最大流,然后加上带费用的边跑费用流
高一的时候做这道题怎么想不到?
注意:maxn代表的不一定是同一个变量的范围
#include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> using namespace std; const int inf=1000000000; const int maxn=5009; int n,m,k; struct Edge{ int from,to,cap,flow,cost; }; vector<int>G[maxn]; vector<Edge>edges; void Addedge(int x,int y,int z,int w){ Edge e; e.from=x;e.to=y;e.cap=z;e.flow=0;e.cost=w; edges.push_back(e); e.from=y;e.to=x;e.cap=0;e.flow=0;e.cost=-w; edges.push_back(e); int c=edges.size(); G[x].push_back(c-2); G[y].push_back(c-1); } int s,t,totn; queue<int>q; int inq[maxn]; int d[maxn]; int p[maxn]; int Spfa(int &nowflow,int &nowcost){ for(int i=1;i<=totn;++i){ d[i]=inf;inq[i]=0; } q.push(s);inq[s]=1;d[s]=0; while(!q.empty()){ int x=q.front();q.pop();inq[x]=0; for(int i=0;i<G[x].size();++i){ Edge e=edges[G[x][i]]; if((e.cap>e.flow)&&(d[x]+e.cost<d[e.to])){ d[e.to]=d[x]+e.cost; p[e.to]=G[x][i]; if(!inq[e.to]){ inq[e.to]=1; q.push(e.to); } } } } if(d[t]==inf)return 0; int x=t,a=inf; while(x!=s){ Edge e=edges[p[x]]; a=min(a,e.cap-e.flow); x=e.from; } nowflow+=a;nowcost+=a*d[t]; x=t; while(x!=s){ edges[p[x]].flow+=a; edges[p[x]^1].flow-=a; x=edges[p[x]].from; } return 1; } int rx[maxn],ry[maxn],rf[maxn],rw[maxn]; int ans,ans2; void Minit(){ ans=ans2=0; for(int i=1;i<=n+1;++i)G[i].clear(); edges.clear(); while(!q.empty())q.pop(); } int main(){ scanf("%d%d%d",&n,&m,&k); Minit(); for(int i=1;i<=m;++i){ scanf("%d%d%d%d",&rx[i],&ry[i],&rf[i],&rw[i]); Addedge(rx[i],ry[i],rf[i],0); } totn=n;s=1;t=n; while(Spfa(ans,ans2)); printf("%d ",ans); Addedge(n+1,1,k,0); totn=n+1;s=n+1; for(int i=1;i<=m;++i){ Addedge(rx[i],ry[i],inf,rw[i]); } while(Spfa(ans,ans2)); printf("%d\n",ans2); return 0; }