P1251 餐巾计划问题 费用流

https://www.luogu.org/problemnew/show/P1251

题意

有一家酒店，酒店每天需要ri张桌布，桌布可以现买，p元。可以通过快洗店，等m天，f元。可以通过慢洗店，等n天，s元。问满足每天用布需求的最小费用

思路

这道题拆点是要的，把一天拆成早上和晚上。比较精彩的是，把每天需要用ri张桌布分开来看，“早上需要有ri张脏布”，“晚上有ri张脏布”。翻译过来就是，早上向终点连ri容量的边，源点向晚上连ri容量的边。
然后又是三种情况的讨论，1）现买，源点向早上连费用为p的边。2）快洗店，晚上向+m天连费用为f的边。3）慢洗店，晚上向+n天连费用为s的边。
最后还要注意，由于可以留下晚上的脏布，所以每个晚上向下一个晚上连边。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

/*
        
⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼('ㅅ')
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ﾚ　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}

/*-----------------------showtime----------------------*/


            const int maxn = 2009;
            struct E{
                int v,w,cost;
                int nxt;
            }edge[10*maxn*maxn];
            int n,gtot = 0;
            int head[10*maxn]; 
            void addedge(int u,int v,int w,int cost){
                edge[gtot].v = v;
                edge[gtot].w = w;
                edge[gtot].cost = cost;
                edge[gtot].nxt = head[u];
                head[u] = gtot ++;

                edge[gtot].v = u;
                edge[gtot].w = 0;
                edge[gtot].cost = -1*cost;
                edge[gtot].nxt = head[v];
                head[v] = gtot++;
            }
            int vis[maxn*10],dis[maxn*10],pre[maxn*10],path[maxn*10];
            bool spfa(int s,int t){
                memset(vis, 0, sizeof(vis));
                memset(dis, inf, sizeof(dis));
                memset(pre, -1, sizeof(pre));

                queue<int>que;  que.push(s);    vis[s] = 1;
                dis[s] = 0;
                while(!que.empty()){
                    int u = que.front(); que.pop(); vis[u] = 0;
                    for(int i=head[u]; ~i; i = edge[i].nxt){
                        int v = edge[i].v, w = edge[i].w, cost = edge[i].cost;
                        if(w > 0 && dis[v] > dis[u] + cost){
                            dis[v] = dis[u] + cost;
                            pre[v] = u; path[v] = i;
                            if(vis[v] == 0){
                                que.push(v);
                                vis[v] = 1;
                            }
                        }
                    }
                }
                return pre[t] != -1;
            }

            ll solve(int s,int t){
                ll flow = 0, cost = 0;
                while(spfa(s,t)){
                    int f = inf;
                    for(int i=t; i!=s; i = pre[i]){
                        f = min(f, edge[path[i]].w);
                    }
                    flow += f;
                    cost += 1ll*f * dis[t];
                    // cout<<f<<" "<<dis[t]<<endl;
                    for(int i=t; i!=s; i = pre[i]){
                        edge[path[i]].w -= f;
                        edge[path[i] ^ 1].w += f;
                    }
                }
                return cost;
            }

int main(){
            memset(head, -1, sizeof(head));
            scanf("%d", &n);
            int s = 0, t = n+n+1;
            for(int i=1; i<=n; i++) {
                int x;
                scanf("%d", &x);
                addedge(s, i+n, x, 0);
                addedge(i, t, x, 0);
            }
            int p,m,ff,nn,ss;
            scanf("%d%d%d%d%d", &p, &m, &ff, &nn, &ss);
            for(int i=1; i<=n; i++) addedge(s, i, inf, p);
        
            for(int i=1; i + m <=n; i++) addedge(i+n, i+m, inf, ff);
            for(int i=1; i + nn<=n; i++) addedge(i+n, i+nn, inf, ss);
        
            for(int i=1; i<n; i++) addedge(i+n, i+n+1, inf, 0);
            printf("%lld
", solve(s, t));
            return 0;
}