luogu P1251 餐巾计划问题 最小费用最大流

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=200010;
#define int long long
bool vis[maxn];
int n,m,s,t,x,y,z,f,cost[maxn],pre[maxn],last[maxn],flow[maxn],maxflow,mincost;
//cost最小花费;pre每个点的前驱；last每个点的所连的前一条边；flow源点到此处的流量
//maxflow 最大流量
//mincost 最大流量的情况下的最小花费
struct Edge
{
    int to,next,flow,cost;//flow流量 cost花费
} edge[maxn];
int head[maxn],num_edge;
queue <int> q;
int t1,m1,t2,m2;
#define INF 2147483647
void add(int from,int to,int flow,int cost)
{
    edge[++num_edge].next=head[from];
    edge[num_edge].to=to;
    edge[num_edge].flow=flow;
    edge[num_edge].cost=cost;
    head[from]=num_edge;

    edge[++num_edge].next=head[to];
    edge[num_edge].to=from;
    edge[num_edge].flow=0;
    edge[num_edge].cost=-cost;
    head[to]=num_edge;
}
bool spfa(int s,int t)
{
    memset(cost,0x7f,sizeof(cost));
    memset(flow,0x7f,sizeof(flow));
    memset(vis,0,sizeof(vis));
    q.push(s);
    vis[s]=1;
    cost[s]=0;
    pre[t]=-1;
    while (!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=0;
        for (int i=head[now]; i!=-1; i=edge[i].next)
        {
            if (edge[i].flow>0 && cost[edge[i].to]>cost[now]+edge[i].cost)//正边
            {
                cost[edge[i].to]=cost[now]+edge[i].cost;
                pre[edge[i].to]=now;
                last[edge[i].to]=i;
                flow[edge[i].to]=min(flow[now],edge[i].flow);//
                if (!vis[edge[i].to])
                {
                    vis[edge[i].to]=1;
                    q.push(edge[i].to);
                }
            }
        }
    }
    return pre[t]!=-1;
}

void MCMF()
{
    while (spfa(s,t))
    {
        int now=t;
        maxflow+=flow[t];
        mincost+=flow[t]*cost[t];
        while (now!=s)
        {
            //从源点一直回溯到汇点
            edge[last[now]].flow-=flow[t];//flow和cost容易搞混
            edge[last[now]^1].flow+=flow[t];
            now=pre[now];
        }
    }
}
signed main()
{
    memset(head,-1,sizeof(head));
    num_edge=-1;
    cin>>n;
    s=0,t=n*2+1;
    int x;
    for (int i=1; i<=n; i++)
    {
        cin>>x;//拆点 
        add(s,i,x,0);//每天晚上从起点获得x条脏餐巾 
        add(i+n,t,x,0);;//每天白天,向汇点提供x条干净的餐巾,流满时表示第i天的餐巾够用 
    }
    scanf("%d %d %d %d %d",&m,&t1,&m1,&t2,&m2);
    for(int i=1; i<=n; i++)
    {
        if(i+1<=n) 
            add(i,i+1,INF,0);//每天晚上可以将脏餐巾留到第二天晚上
        if(i+t1<=n) add(i,i+n+t1,INF,m1);//每天晚上可以送去快洗部,在地i+t1天早上收到餐巾
        if(i+t2<=n) add(i,i+n+t2,INF,m2);//每天晚上可以送去慢洗部,在地i+t2天早上收到餐巾
        add(s,i+n,INF,m);//每天早上可以购买餐巾
    }
    MCMF();
    cout<<mincost<<endl;
    return 0;
}