ss是源点，代表餐巾卖家，tt是汇点，代表记账收钱者。
记p(i)是i天早晨的可用毛巾数，q(i)是i天完了的废毛巾数。
建图见注释
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
typedef long long ll;
int n, mmm, nnn, ss, tt, hea[4005], cnt, pre[4005];
ll r[2005], ppp, fff, sss, minCost, dis[4005];
const ll oo=0x3f3f3f3f3f3f3f3f;
bool vis[4005];
queue<int> d;
struct Edge{
int too, nxt;
ll val, cst;
}edge[25005];
inline int p(int x){
return x;
}
inline int q(int x){
return x+n;
}
void add_edge(int fro, int too, ll val, ll cst){
edge[cnt].nxt = hea[fro];
edge[cnt].too = too;
edge[cnt].val = val;
edge[cnt].cst = cst;
hea[fro] = cnt++;
}
void addEdge(int fro, int too, ll val, ll cst){
add_edge(fro, too, val, cst);
add_edge(too, fro, 0, -cst);
}
bool spfa(){
memset(dis, 0x3f, sizeof(dis));
memset(pre, -1, sizeof(pre));
dis[ss] = 0;
d.push(ss);
vis[ss] = true;
while(!d.empty()){
int x=d.front();
d.pop();
vis[x] = false;
for(int i=hea[x]; i!=-1; i=edge[i].nxt){
int t=edge[i].too;
if(dis[t]>dis[x]+edge[i].cst && edge[i].val>0){
dis[t] = dis[x] + edge[i].cst;
pre[t] = i;
if(!vis[t]){
vis[t] = true;
d.push(t);
}
}
}
}
return dis[tt]!=oo;
}
void dinic(){
while(spfa()){
ll tmp=oo;
for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too])
tmp = min(tmp, edge[i].val);
for(int i=pre[tt]; i!=-1; i=pre[edge[i^1].too]){
edge[i].val -= tmp;
edge[i^1].val += tmp;
minCost += tmp * edge[i].cst;
}
}
}
int main(){
memset(hea, -1, sizeof(hea));
cin>>n;
for(int i=1; i<=n; i++) scanf("%lld", &r[i]);
cin>>ppp>>mmm>>fff>>nnn>>sss;
ss = 0; tt = n + n + 1;
for(int i=1; i<=n; i++){
addEdge(ss, p(i), r[i], ppp);//买毛巾
addEdge(ss, q(i), r[i], 0);//每天完了都要用掉这么多毛巾
addEdge(p(i), tt, r[i], 0);//对每天的新毛巾记账
if(i+mmm<=n) addEdge(q(i), p(i+mmm), oo, fff);//快洗
if(i+nnn<=n) addEdge(q(i), p(i+nnn), oo, sss);//慢洗
if(i<n) addEdge(q(i), q(i+1), oo, 0);//废毛巾滞留
//注意没有必要新毛巾滞留，因为刚好有新毛巾总比滞留新毛巾好
}
dinic();
cout<<minCost<<endl;
return 0;
}