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| \documentclass{article}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{xeCJK}
\usepackage{multicol}
\usepackage{tocloft}
\usepackage{xcolor}
\usepackage{tikz}
\usetikzlibrary{decorations.pathreplacing, arrows.meta}
\usepackage[cachedir=_minted-main]{minted}
\usepackage[a4paper, left=2cm, right=2cm, top=2.5cm, bottom=2.5cm]{geometry}
\setminted{
breaklines=true,
breakanywhere=true,
fontsize=\small,
frame=single,
bgcolor=gray!5,
tabsize=2,
}
\renewcommand{\cftsecleader}{\cftdotfill{\cftdotsep}}
\usepackage[colorlinks=true, linkcolor=black, anchorcolor=black, citecolor=black, filecolor=black, menucolor=black, runcolor=black, urlcolor=blue]{hyperref}
\definecolor{seg1}{RGB}{50,120,200}
\definecolor{seg2}{RGB}{220,80,60}
\definecolor{seg3}{RGB}{0,160,80}
\definecolor{seg4}{RGB}{140,80,200}
\definecolor{highlight}{RGB}{255,160,0}
\title{\textbf{P16230 [蓝桥杯 2026 省 A] 综合应变指标} \\ 题解}
\author{}
\date{}
\begin{document}
\maketitle
\section{题意概述}
给定长度为 $N$ 的整数序列 $A$，选取三个分割点 $i, j, k$（满足 $1 \le i < j < k < N$）将序列划分为四个非空连续子段，最大化四段元素和的绝对值之和：
$$|A_1 + \cdots + A_i| + |A_{i+1} + \cdots + A_j| + |A_{j+1} + \cdots + A_k| + |A_{k+1} + \cdots + A_N|$$
\section{第一层思考：前缀和改写目标函数}
定义前缀和 $P_m = \sum_{t=1}^{m} A_t$（$P_0 = 0$）。四段的元素和分别为 $P_i$、$P_j - P_i$、$P_k - P_j$、$P_N - P_k$。目标函数改写为：
$$f(i, j, k) = |P_i| + |P_j - P_i| + |P_k - P_j| + |P_N - P_k|$$
这一步把"子段求和"全部消解成了前缀和之差的绝对值，后续只需操作 $P$ 数组。
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=0.75, every node/.style={font=\small}]
\foreach \idx in {1,...,10} {
\fill[gray!12] (\idx-0.45, -0.4) rectangle (\idx+0.45, 0.4);
\draw[gray!50] (\idx-0.45, -0.4) rectangle (\idx+0.45, 0.4);
\node at (\idx, 0) {$A_{\idx}$};
}
\draw[seg1, line width=3pt] (0.45, -0.65) -- (3.55, -0.65);
\node[seg1, font=\bfseries] at (2, -1.15) {第 1 段};
\node[seg1] at (2, -1.65) {和 $= P_i$};
\draw[seg2, line width=3pt] (3.55, -0.65) -- (6.55, -0.65);
\node[seg2, font=\bfseries] at (5, -1.15) {第 2 段};
\node[seg2] at (5, -1.65) {和 $= P_j - P_i$};
\draw[seg3, line width=3pt] (6.55, -0.65) -- (8.55, -0.65);
\node[seg3, font=\bfseries] at (7.5, -1.15) {第 3 段};
\node[seg3] at (7.5, -1.65) {和 $= P_k - P_j$};
\draw[seg4, line width=3pt] (8.55, -0.65) -- (10.55, -0.65);
\node[seg4, font=\bfseries] at (9.5, -1.15) {第 4 段};
\node[seg4] at (9.5, -1.65) {和 $= P_N - P_k$};
\draw[thick, dashed] (3.55, 0.6) -- (3.55, -0.55);
\node[above] at (3.55, 0.6) {$i$};
\draw[thick, dashed] (6.55, 0.6) -- (6.55, -0.55);
\node[above] at (6.55, 0.6) {$j$};
\draw[thick, dashed] (8.55, 0.6) -- (8.55, -0.55);
\node[above] at (8.55, 0.6) {$k$};
\end{tikzpicture}
\caption{
将序列 $A$ 在分割点 $i, j, k$ 处划分为四段。每段的元素和均可用前缀和 $P$ 简洁表示。
}
\label{fig:partition}
\end{figure}
直接枚举三个分割点的时间复杂度为 $O(N^3)$，对 $N = 10^5$ 不可接受。能否把三重循环逐层拆解，每层都只做线性扫描？
\section{第二层思考：$O(N^2)$ 的朴素代码}
问题本身就是分段的，自然地逐段转移。定义四层 DP：\texttt{dp1[i]} 表示只切出第一段、该段以位置 $i$ 结尾时的最大贡献；\texttt{dp2[i]} 表示前两段的最大贡献，第二段以 $i$ 结尾；\texttt{dp3}、\texttt{dp4} 同理。答案为 \texttt{dp4[N]}。
第一层直接算：
\begin{minted}{cpp}
dp1[i] = abs(preA[i]);
\end{minted}
从 \texttt{dp1} 到 \texttt{dp2}、从 \texttt{dp2} 到 \texttt{dp3}、从 \texttt{dp3} 到 \texttt{dp4}，转移结构完全一样，写成一个通用函数：
\begin{minted}{cpp}
void excute_dp(const vector<ll> &dp1, vector<ll> &dp2) {
for (int i = 1; i <= N; ++i) {
for (int j = 1; j <= i; ++j) { // ← 内层循环
dp2[i] = max(dp2[i], dp1[j] + abs(preA[i] - preA[j]));
}
}
}
\end{minted}
调用三次 \texttt{excute\_dp}，每次 $O(N^2)$。瓶颈就在内层 \texttt{for(j)} 循环——能不能干掉它？
\section{第三层思考：干掉内层循环}
盯着内层循环体看：
\begin{minted}{cpp}
dp2[i] = max(dp2[i], dp1[j] + abs(preA[i] - preA[j]));
\end{minted}
\texttt{abs} 只有两种展开。代入后，把只跟 \texttt{j} 有关的项括起来：
\begin{minted}{cpp}
dp1[j] + preA[i] - preA[j] = (dp1[j] - preA[j]) + preA[i]
// ^^^^^^^^^^^^^^^^^^ ^^^^^^^^
// 只跟 j 有关 只跟 i 有关
dp1[j] + preA[j] - preA[i] = (dp1[j] + preA[j]) - preA[i]
// ^^^^^^^^^^^^^^^^^^ ^^^^^^^^
// 只跟 j 有关 只跟 i 有关
\end{minted}
所以内层循环等价于：
\begin{minted}{cpp}
dp2[i] = max(
max_over_j(dp1[j] - preA[j]) + preA[i],
max_over_j(dp1[j] + preA[j]) - preA[i]
);
\end{minted}
\texttt{max\_over\_j(dp1[j] - preA[j])} 和 \texttt{max\_over\_j(dp1[j] + preA[j])} 不依赖 \texttt{i}！而且随着 \texttt{i} 递增，\texttt{j} 的候选范围只增不减（$j \le i$）。用两个变量滚动维护前缀最大值就行了，内层循环彻底消失：
\begin{minted}{cpp}
void excute_dp(const vector<ll> &from, vector<ll> &to) {
ll Mplus = -INF, Mminus = -INF;
for (int i = 1; i <= N; ++i) {
Mplus = max(Mplus, from[i] + preA[i]); // 维护 max(dp1[j]+preA[j])
Mminus = max(Mminus, from[i] - preA[i]); // 维护 max(dp1[j]-preA[j])
to[i] = max(Mminus + preA[i], Mplus - preA[i]);
}
}
\end{minted}
双重循环 $\to$ 单层循环，$O(N^2) \to O(N)$。调用三次，总计 $O(N)$。
\subsection{图像理解}
上面的优化可以用图~\ref{fig:v-shape} 直观看懂。
对每个候选 \texttt{j}，\texttt{dp1[j] + abs(preA[i] - preA[j])} 是关于 \texttt{preA[i]} 的一条 V 形曲线——谷底在 \texttt{preA[j]}，谷底高 \texttt{dp1[j]}，两臂斜率 $\pm 1$。内层循环就是对所有 V 形取最大值（上包络）。
关键：所有 V 形的斜率全一样（$\pm 1$），不同的只是截距。同斜率的平行线取最大值，只有截距最大的一条存活。所以上包络只由两条直线决定——截距就是代码里的 \texttt{Mminus} 和 \texttt{Mplus}。
\texttt{i} 每递增一步，候选集多出一条新 V 形。新 V 形的截距和当前 \texttt{Mminus}、\texttt{Mplus} 取一次 \texttt{max} 就完成更新——这就是循环体里那两行 \texttt{max} 在干的事。
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=0.85, every node/.style={font=\small}]
\begin{scope}
\clip (-0.5, -0.5) rectangle (8.2, 7.5);
\draw[gray!40, line width=0.5pt, densely dashed] (0, 4) -- (2, 2);
\draw[gray!40, line width=0.5pt, densely dashed] (0, 5.5) -- (4, 1.5) -- (7.5, 5);
\draw[gray!40, line width=0.5pt, densely dashed] (6, 1) -- (8, 3);
\draw[seg1, line width=2.5pt] (2, 2) -- (7.5, 7.5);
\draw[seg3, line width=2.5pt] (0, 7) -- (6, 1);
\draw[highlight, line width=3.5pt, opacity=0.45] (0, 7) -- (3.5, 3.5) -- (7.5, 7.5);
\end{scope}
\fill[seg1] (2, 2) circle (3pt);
\fill[seg2] (4, 1.5) circle (3pt);
\fill[seg3] (6, 1) circle (3pt);
\fill[highlight!80!black] (3.5, 3.5) circle (3pt);
\draw[seg1!40, densely dotted, thin] (0, 2) -- (2, 2);
\draw[seg2!40, densely dotted, thin] (0, 1.5) -- (4, 1.5);
\draw[seg3!40, densely dotted, thin] (0, 1) -- (6, 1);
\draw[-stealth, thick] (-0.5, 0) -- (8.2, 0) node[right] {\texttt{preA[i]}};
\draw[-stealth, thick] (0, -0.5) -- (0, 7.8);
\foreach \x/\col/\lab in {2/seg1/1, 4/seg2/2, 6/seg3/3} {
\draw[\col] (\x, -0.08) -- (\x, 0.08);
\node[\col, below, font=\footnotesize] at (\x, -0.2) {\texttt{preA[$j_\lab$]}};
}
\node[seg1, left, font=\footnotesize] at (-0.15, 2) {\texttt{dp1[$j_1$]}};
\node[seg2, left, font=\footnotesize] at (-0.15, 1.5) {\texttt{dp1[$j_2$]}};
\node[seg3, left, font=\footnotesize] at (-0.15, 1) {\texttt{dp1[$j_3$]}};
\draw[-stealth, seg1!80!black, thick] (7.2, 5.5) -- (6.2, 6.2);
\node[seg1!80!black, font=\footnotesize, align=left, right] at (7.2, 5.0)
{斜率 $+1$ 冠军\\[-1pt] 截距 $=$ \texttt{Mminus}};
\draw[-stealth, seg3!80!black, thick] (0.6, 7.5) -- (1.2, 5.8);
\node[seg3!80!black, font=\footnotesize, align=left, right] at (0.6, 7.7)
{斜率 $-1$ 冠军，截距 $=$ \texttt{Mplus}};
\draw[gray!40, line width=0.5pt, densely dashed] (4.5, -1.2) -- (5.3, -1.2);
\node[gray!60, font=\footnotesize, right] at (5.4, -1.2) {被淘汰的臂};
\draw[highlight, line width=3pt, opacity=0.45] (4.5, -1.7) -- (5.3, -1.7);
\node[highlight!80!black, font=\footnotesize, right] at (5.4, -1.7) {上包络 $=$ \texttt{dp2[i]}};
\end{tikzpicture}
\caption{
每个候选 \texttt{j} 产生一条 V 形（谷底高 \texttt{dp1[j]}，位于 \texttt{preA[j]}）。所有 V 形斜率一样（$\pm 1$），上包络只由截距最大的两条臂决定——截距就是代码里的 \texttt{Mminus} 和 \texttt{Mplus}。其余臂（灰色虚线）全被淘汰。
}
\label{fig:v-shape}
\end{figure}
\section{代码实现}
\begin{minted}{cpp}
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
static constexpr ll INF = (1ll << 61) - 1;
struct Solve {
ll N;
vector<ll> A, preA;
vector<ll> dp1, dp2, dp3, dp4;
// 核心：优化后的转移，O(N)
void excute_dp(const vector<ll> &from, vector<ll> &to) {
ll Mplus = -INF, Mminus = -INF;
for (int i = 1; i <= N; ++i) {
Mplus = max(Mplus, from[i] + preA[i]);
Mminus = max(Mminus, from[i] - preA[i]);
to[i] = max(Mminus + preA[i], Mplus - preA[i]);
}
}
Solve() {
cin >> N;
A.resize(N + 2);
for (int i = 1; i <= N; ++i)
cin >> A[i];
preA.resize(N + 2);
partial_sum(A.begin(), A.end(), preA.begin());
dp1.resize(N + 2, -INF);
for (int i = 1; i <= N; ++i)
dp1[i] = abs(preA[i]);
dp2.resize(N + 2, -INF);
dp3.resize(N + 2, -INF);
dp4.resize(N + 2, -INF);
excute_dp(dp1, dp2);
excute_dp(dp2, dp3);
excute_dp(dp3, dp4);
cout << dp4[N] << "\n";
}
};
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
Solve solve;
return 0;
}
\end{minted}
\section{复杂度分析}
\begin{itemize}
\item \textbf{时间复杂度}：$O(N)$。\texttt{excute\_dp} 单次 $O(N)$，调用三次，总计 $O(N)$。
\item \textbf{空间复杂度}：$O(N)$。四个 DP 数组加前缀和数组。
\end{itemize}
\end{document}
|
