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>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#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 '\n'

#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\n", solve(s, t));
    return 0;
}