Solution -「NEERC 2016」Delight for a Cat 的一个尝试

\(\mathscr{Description}\)

  Link.

  给定 \(n,k,m_s,m_e\) 和两个长为 \(n\) 的序列 \(\{s\},\{e\}\), 选择一个 \(S\subseteq[1,n]\cap\mathbb N\), 满足 \(\forall l\in[1,n-k+1],~\sum_{i=l}^{l+k-1}[i\in S]\in[m_s,k-m_e]\), 并最大化 \(\sum_{i=1}^n([i\in S]s_i+[i\notin S]e_i)\).

  \(n\le10^3\).

\(\mathscr{Solution}\)

  这里一个机械化方法的尝试, 没写代码是因为不可告人的原因. (

  我们想求

\[\begin{array}{rll} \max & \displaystyle \sum_{i=1}^ne_i+\sum_{i=1}^n(s_i-e_i)(x_i-x_{i-1})\\ \text{s.t.} & x_0=0\\ & x_i\ge x_{i-1} & (\forall i>0)\\ & x_{i-1}-x_{i}\ge-1 & (\forall i>0) \\ & x_i-x_{i-k}\ge m_s & (\forall i\ge k)\\ & x_{i-k}-x_i\ge m_e-k & (\forall i\ge k) \end{array}\tag1 \]

  先忽略 \(x_0=0\) 这个约束. 从现在起, 我们停止思考, 直接去套这个 trick. 通过足够大的代价 \(I\) 间接限定 \(\ge\):

\[\begin{aligned} (1)\Leftrightarrow \max\quad & \sum_{i=1}^n e_i+\sum_{i=1}^n(s_i-s_{i+1}-e_i+e_{i+1})x_i\\ &-\sum_{i=1}^n\max\{x_{i-1}-x_i,0\}I\\ &-\sum_{i=1}^n\max\{x_i-x_{i-1}-1,0\}I\\ &-\sum_{i=k}^n\max\{m_s+x_{i-k}-x_i,0\}I\\ &-\sum_{i=k}^n\max\{m_e-k+x_i-x_{i-k},0\}I. \end{aligned} \]

注意到这个形式满足模型

\[\min\quad \sum_{u}b_ux_u+\sum_{\lang v,u\rang}c_{uv}\max\{0,x_v-x_u-w_{uv}\}, \]

因此可以直接翻译为费用流建图. 先取个负把 \(\max\) 变成 \(\min\), 令 \(b_i=s_{i+1}-s_i+e_i-e_{i+1}\), 流网络 \(G=(V,E)\) 建图如下:

\[\begin{aligned} V &= \{S,T\}\cup\{1,2,\dots,n\},\\ E &= \{\lang S,i,b_i,0\rang\mid b_i>0\} \cup \{\lang u,T,-b_i,0\rang\mid b_i<0\}\\ &\cup \{\lang i,i-1,I,0\rang\}\cup\{\lang i-1,i,I,1\rang\}\\ &\cup \{\lang i,i-k,I,-m_s\rang\}\cup\{\lang i-k,i,I,k-m_e\rang\}. \end{aligned} \]

设最小费用为 \(c\), 对应的解集 \(\{x_{0..n}\}\). 注意此时可能有 \(x_0\neq0\), 所以需要令 \(x_i\gets x_i-x_0\) 并得到实际的最小费用 \(c'\), 则 \(\sum_{i=1}^ne_i+c'\) 就是答案, \(\{x_{0..n}\}\) 的差分就是方案. 复杂度为 \(\mathcal O(\text{Dinic}(n,5n))\).

  不过 ... 求这个解集 \(\{x_{0..n}\}\) 并不太容易. 神 crashed 称可以将残量网络上的 \(f_{uv}\) 对偶回原线规, 然后根据互松弛定理得到关于 \(x\) 的取等条件, 最后建差分约束解出 \(x\). 现在你知道我没有写代码的原因啦!

\(\mathscr{Code}\)

  放一下传统建图的代码, 直接读建图部分不难理解. 复杂度是一样的.

/*+Rainybunny+*/

#include <bits/stdc++.h>

#define rep(i, l, r) for (int i = l, rep##i = r; i <= rep##i; ++i)
#define per(i, r, l) for (int i = r, per##i = l; i >= per##i; --i)

typedef long long LL;
typedef std::pair<int, LL> PIL;
typedef std::pair<LL, int> PLI;
#define fi first
#define se second

const int MAXN = 1e3;
int n, k, ms, me, s[MAXN + 5], e[MAXN + 5];

namespace FG {

// #define POSITIVE_FLAG // determined if spfa is necessary.

const int MAXND = MAXN + 1, MAXEG = MAXN * 2 + 1;
const LL LINF = 1ll << 60;
int S, T, ecnt = 1, head[MAXND + 5], curh[MAXND + 5];
LL dis[MAXND + 5], hgt[MAXND + 5];
struct Edge { int to, flw, cst, nxt; } graph[MAXEG * 2 + 5];
bool instk[MAXND + 5];

inline void link(const int s, const int t, const int f, const int c) {
    // printf("%d %d %d,%d\n", s, t, f, c);
    graph[++ecnt] = { t, f, c, head[s] }, head[s] = ecnt;
    graph[++ecnt] = { s, 0, -c, head[t] }, head[t] = ecnt;
}

#ifndef POSITIVE_FLAG
inline bool spfa() {
    static std::queue<int> que;
    static bool inq[MAXND + 5];
    rep (i, S, T) dis[i] = LINF;
    dis[S] = 0, que.push(S);
    while (!que.empty()) {
        int u = que.front(); que.pop(), inq[u] = false;
        for (int i = head[u], v; i; i = graph[i].nxt) {
            if (graph[i].flw && dis[v = graph[i].to] > dis[u] + graph[i].cst) {
                dis[v] = dis[u] + graph[i].cst;
                if (!inq[v]) inq[v] = true, que.push(v);
            }
        }
    }
    return dis[T] != LINF;
}
#endif

inline bool dijkstra() {
    static std::priority_queue<PLI, std::vector<PLI>, std::greater<PLI>> heap;
    rep (i, S, T) hgt[i] += dis[i], dis[i] = LINF;
    heap.emplace(dis[S] = 0, S);
    while (!heap.empty()) {
        PLI p(heap.top()); heap.pop();
        if (p.fi != dis[p.se]) continue;
        for (int i = head[p.se], v; i; i = graph[i].nxt) {
            LL d = p.fi + graph[i].cst - hgt[v = graph[i].to] + hgt[p.se];
            if (graph[i].flw && d < dis[v]) heap.emplace(dis[v] = d, v);
        }
    }
    return dis[T] != LINF;
}

inline PIL augment(const int u, int iflw) {
    if (u == T) return { iflw, 0 };
    PIL ret(0, 0); instk[u] = true;
    for (int &i = curh[u], v; i; i = graph[i].nxt) {
        if (graph[i].flw && !instk[v = graph[i].to]
          && dis[v] == dis[u] + graph[i].cst - hgt[v] + hgt[u]) {
            PIL t(augment(v, std::min(iflw, graph[i].flw)));
            graph[i].flw -= t.fi, graph[i ^ 1].flw += t.fi;
            ret.fi += t.fi, iflw -= t.fi;
            ret.se += t.se + 1ll * graph[i].cst * t.fi;
            if (!iflw) break;
        }
    }
    if (ret.fi) instk[u] = false;
    return ret;
}

inline PIL dinic() {
    PIL ret(0, 0);
#ifdef POSITIVE_FLAG
    while (dijkstra()) {
#else
    for (spfa(); dijkstra();) {
#endif
        rep (i, S, T) curh[i] = head[i], instk[i] = false;
        PIL t(augment(S, 0x3f3f3f3f));
        ret.fi += t.fi, ret.se += t.se;
    }
    return ret;
}

} // namespace FG.

int main() {
    scanf("%d %d %d %d", &n, &k, &ms, &me);
    rep (i, 1, n) scanf("%d", &s[i]);
    rep (i, 1, n) scanf("%d", &e[i]);
 
    FG::S = 0, FG::T = n + 1;
    rep (i, 1, n) FG::link(i, std::min(i + k, n + 1), 1, s[i] - e[i]);
    rep (i, 1, k - 1) FG::link(i, i + 1, k - ms, 0);
    rep (i, k, n) FG::link(i, i + 1, k - ms - me, 0);
    FG::link(FG::S, 1, k - ms, 0);

    LL ans = 0;
    rep (i, 1, n) ans += s[i];
    printf("%lld\n", ans - FG::dinic().se);
    rep (i, 1, n) putchar("ES"[FG::graph[i << 1].flw]);
    putchar('\n');
    return 0;
}