[ZJOI2010]网络扩容

题目

[ZJOI2010]网络扩容
(A：)报告，发现一道水题
(B：)切掉切掉

做法

考虑做第二问，(ufrac{~~f~~}{~~0~~}v,ufrac{~~K~~}{~~c~~}v)，然后跑最大流最小费用就好了

My complete code

#include<bits/stdc++.h>
using namespace std;
typedef int LL;
const LL maxn=30009,inf=0x3f3f3f3f;
LL n,m,S,T,K,ans;
LL U[maxn],V[maxn],F[maxn],W[maxn];
struct E{
	struct node{
	    LL to,next,f,c;
    }dis[maxn];
    LL num;
    LL head[maxn],lev[maxn];
    bool visit[maxn];
    inline void Init(){
    	memset(head,-1,sizeof(head)), num=-1;
	}
	
    inline void Add(LL u,LL v,LL f,LL c=0){
    	dis[++num]=(node){v,head[u],f,c}, head[u]=num;
	}
	inline bool Bfs(){
		queue<LL> que; 
		memset(lev,0,sizeof(lev)), memset(visit,false,sizeof(visit));
		que.push(S); visit[S]=true; lev[S]=0;
		while(que.size()){
			LL u(que.front()); que.pop();
			for(LL i=head[u];~i;i=dis[i].next){
				LL v(dis[i].to);
				if(!visit[v] && dis[i].f)
				    lev[v]=lev[u]+1,visit[v]=true,que.push(v);
			}
		}return lev[T];
	}
	LL Dfs(LL u,LL f){
		if(u==T) return f;
		LL tmp(f);
		for(LL i=head[u];~i;i=dis[i].next){
			LL v(dis[i].to),now;
			if(dis[i].f &&lev[v]==lev[u]+1){
				if(now=Dfs(v,min(tmp,dis[i].f))){
					dis[i].f-=now, dis[i^1].f+=now;
					tmp-=now;
					if(!tmp) break;
				}else lev[v]=-1;
			}
		}return f-tmp;
	}
	inline LL Dinic(){
		LL ret(0);
		while(Bfs()) ret+=Dfs(S,inf);
		return ret;
	}
	
	
	LL pre_d[maxn],pre_v[maxn],cost[maxn],flow[maxn];
	inline bool EK(){
		memset(pre_d,0,sizeof(pre_d)), memset(pre_v,0,sizeof(pre_v)), memset(cost,inf,sizeof(cost));
		memset(flow,0,sizeof(flow));
		queue<LL> que;
		que.push(S);
		cost[S]=0, flow[S]=inf;
		while(que.size()){
			LL u(que.front()); que.pop();
			for(LL i=head[u];~i;i=dis[i].next){
				LL v(dis[i].to);
				if(cost[v]>cost[u]+dis[i].c && dis[i].f){
					flow[v]=min(flow[u],dis[i].f);
					cost[v]=cost[u]+dis[i].c;
					pre_d[v]=i, pre_v[v]=u;
					que.push(v);
				}
			}
		}return pre_v[T];
	}
	inline LL Solve(){
		LL ret(0);
		while(EK()){
			ret+=flow[T]*cost[T];
			for(LL i=T;i!=S;i=pre_v[i]){
				dis[pre_d[i]].f-=flow[T],
				dis[pre_d[i]^1].f+=flow[T];
			}
		}return ret;
	}
}G1,G2;
int main(){
	cin>>n>>m>>K;
	G1.Init();
	S=1, T=n;
	for(LL i=1;i<=m;++i)
		cin>>U[i]>>V[i]>>F[i]>>W[i],
		G1.Add(U[i],V[i],F[i]), G1.Add(V[i],U[i],0);
	ans=G1.Dinic();
	printf("%d ",ans);
	
	G2.Init();
	S=n+1,T=n+2;
	G2.Add(S,1,ans+K,0),G2.Add(1,S,0,0);
	G2.Add(n,T,ans+K,0),G2.Add(T,n,0,0);
	for(LL i=1;i<=m;++i){
		G2.Add(U[i],V[i],F[i],0), G2.Add(V[i],U[i],0,0);
		G2.Add(U[i],V[i],K,W[i]), G2.Add(V[i],U[i],0,-W[i]);
	}
	ans=G2.Solve();
	printf("%d",ans);
	return 0;
}