把每天拆成两个点，需要用的餐巾数和用了的餐巾数，分别连源和汇，流量day[i]，费用0
再从源点每天用的餐巾点，费用为p，流量无穷。
每天需要用的餐巾连下一天需要用的（没洗的），流量无穷，费用0；
每天用完的连n和m天后的用了的餐巾，费用f,s，流量无穷
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int inf = 100000000;
const static int V = 2010;
const static int E = 400000;
int sink, source, dis[V], pre[V];
bool vst[V];
int day[V];
struct Edge
{
int v, flow, cost;
Edge *next;
}pool[E], *adj[V],*pree[V],*pp;
void create(int u, int v, int flow, int cost)
{
pp->flow = flow;
pp->v = v;
pp->cost = cost;
pp->next = adj[u];
adj[u] = pp++;
}
void addedge(int u, int v, int flow, int cost)
{
//printf("%d %d %d %d\n",u,v,flow,cost);
create(u,v,flow,cost);
create(v,u,0,-cost);
}
void SPFA()
{
int u;
for (int i = 0; i < V; ++i)
dis[i] = inf;
memset(vst,0,sizeof(vst));
queue<int> q;
q.push(source);
vst[source] = 1;
dis[source] = 0;
while(!q.empty())
{
int v = q.front();
q.pop();
for (Edge *i = adj[v]; i != NULL; i = i->next)
{
if (!i->flow)
continue;
u = i->v;
if (i->cost + dis[v] < dis[u])
{
dis[u] = i->cost + dis[v];
pree[u] = i;
pre[u] = v;
if (!vst[u])
{
q.push(u);
vst[u] = true;
}
}
}
vst[v] = false;
}
}
void init()
{
memset(adj, 0, sizeof (adj));
pp = pool;
}
int Mincost_Maxflow()
{
int flow = 0, cost = 0;
while (true)
{
SPFA();
if (dis[sink] == inf)
break;
int minn = inf;
for (int i = sink; i != source; i = pre[i])
minn = min(minn, pree[i]->flow);
flow += minn;
for (int i = sink; i != source; i = pre[i])
{
pree[i]->flow -= minn;
pool[(pree[i] - pool)^0x1].flow += minn;
cost += minn * pree[i]->cost;
}
}
return cost;
}
int main()
{
int N,p,m,f,n,s;
while(scanf("%d %d %d %d %d %d",&N,&p,&m,&f,&n,&s) == 6)
{
init();
source = 0;
sink = 2*N + 1;
for(int i=1;i<=N;i++)
{
scanf("%d",&day[i]);
addedge(source,2*i-1,day[i],0);
addedge(source,2*i,inf,p);
addedge(2*i,sink,day[i],0);
if(i + 1 <= N)
addedge(2*i - 1,2*(i+1) - 1,inf,0);
if(i + m <=N)
addedge(2*i - 1,2*(i + m),inf,f);
if(i + n <=N)
addedge(2*i - 1,2*(i + n),inf,s);
}
int ff = Mincost_Maxflow();
printf("%d\n",ff);
}
return 0;
}