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P1251 餐巾计划问题

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P1251 餐巾计划问题

题面

题目链接

解法

题目中已经暗示了要用费用流。

使用流模型，流表示餐巾的数量。但是不易处理餐巾干净与否，所以拆点，将一天拆成两个点，分别表示这天拥有的干净餐巾数量和脏的餐巾数量。为方便表述，分别设为$used{i,0}$和$used{i,1}$。

由于每天需要使用$ri$块餐巾，所以让$used{i,0}$连向汇点$T$，费用为$0$，流量为$ri$，表示这天至少需要拥有$r_i$块干净餐巾。同时让源点$S$连向$used{i,1}$，费用为$0$，流量为$r_i$，表示这天过后会多出$r_i$块脏的餐巾。

接下来考虑购买餐巾，显然可以直接从$S$连向$used{i,0}$，费用为$p$，流量为$+\infty$。而使用快洗部则从$used{i,1}$连向$used{i+m,0}$，费用为$f$，流量为$+\infty$。同理，慢洗部$used{i,1}$连向$used{i+n,0}$，费用为$s$，流量为$+\infty$。另一种保留餐巾则$used{i,1}$连向$used_{i+1,1}$，费用为$0$，流量为$+\infty$。

形式化地，定义函数addedge(u, v, cost, flow)，则需要

  1. addedge(used[i][0], T, 0, r[i])
  2. addedge(S, used[i][1], 0, r[i])
  3. addedge(S, used[i][0], p, INF)
  4. if (i + 1 <= N) addedge(used[i][1], used[i + 1][1], 0, INF)
  5. if (i + m <= N) addedge(used[i][1], used[i + m][0], f, INF)
  6. if (i + n <= N) addedge(used[i][1], used[i + n][0], s, INF)

AC代码

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/**
 * @file:           P1251.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P1251
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
const int MAXN = 1e4 + 10;
const int MAXM = 5e4 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct Dinic {
    struct Edge {
        int v, cost, flow;
    } edge[MAXM];
    int tot = 1, cost = 0, flow = 0;
    int head[MAXN], nxt[MAXM], dis[MAXN], cur[MAXN];
    bool vis[MAXN];
    void addedge(int u, int v, int cost, int flow) {
        edge[++tot] = {v, cost, flow};
        nxt[tot] = head[u], head[u] = tot;
        edge[++tot] = {u, -cost, 0};
        nxt[tot] = head[v], head[v] = tot;
    }
    bool spfa(int s, int t) {
        memset(vis, 0, sizeof(vis));
        memset(dis, 0x3f, sizeof(dis));
        queue<int> que;
        que.push(s);
        dis[s] = 0, vis[s] = true;
        while (!que.empty()) {
            int u = que.front();
            que.pop(), vis[u] = false;
            for (int i = head[u]; i; i = nxt[i]) {
                int v = edge[i].v;
                if (edge[i].flow && dis[v] > dis[u] + edge[i].cost) {
                    dis[v] = dis[u] + edge[i].cost;
                    if (!vis[v]) {
                        que.push(v);
                        vis[v] = true;
                    }
                }
            }
        }
        return dis[t] != INF;
    }
    int augment(int u, int t, int mx) {
        if (u == t || mx == 0)
            return mx;
        vis[u] = true;
        int ret = 0;
        for (int &i = cur[u]; i; i = nxt[i]) {
            int v = edge[i].v;
            if (vis[v] || dis[u] + edge[i].cost != dis[v])
                continue;
            int tmp = augment(v, t, min(mx, edge[i].flow));
            mx -= tmp, ret += tmp;
            edge[i].flow -= tmp, edge[i ^ 1].flow += tmp;
            cost += edge[i].cost * tmp;
            if (mx == 0)
                break;
        }
        vis[u] = false;
        return ret;
    }
    void mcmf(int s, int t) {
        while (spfa(s, t)) {
            memcpy(cur, head, sizeof(cur));
            flow += augment(s, t, INF);
        }
    }
};
int n, qp, qm, qf, qn, qs;
int a[MAXN], used[MAXN][2];
Dinic mcmf;
signed main() {
    read(n);
    for (int i = 1; i <= n; ++i)
        read(a[i]);
    read(qp), read(qm), read(qf), read(qn), read(qs);
    int s = 1, t = 2;
    for (int i = 1; i <= n; ++i)
        used[i][0] = i * 2 + 1, used[i][1] = i * 2 + 2;
    for (int i = 1; i <= n; ++i) {
        mcmf.addedge(used[i][0], t, 0, a[i]);
        mcmf.addedge(s, used[i][1], 0, a[i]);
        mcmf.addedge(s, used[i][0], qp, INF);
        if (i + 1 <= n)
            mcmf.addedge(used[i][1], used[i + 1][1], 0, INF);
        if (i + qm <= n)
            mcmf.addedge(used[i][1], used[i + qm][0], qf, INF);
        if (i + qn <= n)
            mcmf.addedge(used[i][1], used[i + qn][0], qs, INF);
    }
    mcmf.mcmf(s, t);
    write(mcmf.cost), putchar('\n');
    return 0;
}