luogu1251 餐巾计划问题

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;
}