2017.10.11 network 网络扩容 思考记录

第一问是网络流。

第二问是费用流，但之前的边依然是可以用的，所以每个点决策：用原来的流量    购买一条流量，扫每条边加进去就可以了

然后限制流出为k

码：

#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;#define N 1005#define M 50005#define inf 10000007int tot=-1,hou[M],yi[M],er[M],san[M],si[M],s,t,xia[N],zhong[M],yuan[M],v[M],co[M],dis[N],daan,n,m,k,i,j,qj[N];bool vis[N];queue<int>q;void jian(int a,int b,int c,int d){++tot;hou[tot]=yuan[a],yuan[a]=tot,zhong[tot]=b,v[tot]=c,co[tot]=d;}void jia(int a,int b,int c,int d){jian(a,b,c,d);jian(b,a,0,-d);}bool bfs(){for(i=1;i<=n;i++){xia[i]=yuan[i];dis[i]=inf;}q.push(s);dis[s]=1;while(!q.empty()){int st=q.front();q.pop();for(i=xia[st];i!=-1;i=hou[i]){int nd=zhong[i];if(dis[nd]!=inf||v[i]==0)continue;dis[nd]=dis[st]+1;q.push(nd);}}return dis[t]<inf;}int dfs(int o,int t,int limit){if(o==t||!limit)return limit;int i,f,flow=0;for(i=xia[o];i!=-1;i=hou[i]){xia[o]=i;int nd=zhong[i];if(dis[nd]==dis[o]+1&&(f=dfs(nd,t,min(limit,v[i])))){limit-=f;flow+=f;v[i]-=f;v[i^1]+=f;if(!limit)break;}}return flow;}int dinic(){int ans=0;while(bfs()){ans+=dfs(s,t,inf);}return ans;}bool spfa(){qj[s]=-1;for(i=1;i<=n;i++){dis[i]=inf;}q.push(s);dis[s]=0;while(!q.empty()){int st=q.front();vis[st]=0;q.pop();for(i=yuan[st];i!=-1;i=hou[i]){int nd=zhong[i];if(v[i]>0&&dis[nd]>dis[st]+co[i]){dis[nd]=dis[st]+co[i];qj[nd]=i^1;if(vis[nd]==0){q.push(nd);vis[nd]=1;}}}}if(dis[t]==inf)return 0;int o,minn=inf;for(o=qj[t];o!=-1;o=qj[zhong[o]]){minn=min(minn,v[o^1]);}for(o=qj[t];o!=-1;o=qj[zhong[o]]){v[o]+=minn;v[o^1]-=minn;daan+=minn*co[o^1];}return 1;}int mcmf(){while(spfa());}int main(){memset(yuan,-1,sizeof(yuan));scanf("%d%d%d",&n,&m,&k);s=1;t=n;for(i=1;i<=m;i++){scanf("%d%d%d%d",&yi[i],&er[i],&san[i],&si[i]);jia(yi[i],er[i],san[i],0);}printf("%d",dinic());for(i=1;i<=m;i++){jia(yi[i],er[i],15,si[i]);}++n;jia(t,n,k,0);t=n;mcmf();printf(" %d",daan);}