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# Slope Trick

Author: Benjamin Qi

### Prerequisites

Slope trick refers to a way to manipulate piecewise linear convex functions. Includes a simple solution to USACO Landscaping.

## Tutorials

Resources
CF

3 problems using this trick

CF

clarifying the above and another example problem

From the latter link (modified):

Slope trick is a way to represent a function that satisfies the following conditions:

• It can be divided into multiple sections, where each section is a linear function (usually) with an integer slope.
• It is a convex/concave function. In other words, the slope of each section is non-decreasing or non-increasing when scanning the function from left to right.

It's generally applicable as a DP optimization.

### Pro Tip

Usually you can come up with a slower (usually $\mathcal{O}(N^2)$) DP first and then optimize it to $\mathcal{O}(N\log N)$ with slope trick.

The rest of this module assumes that you know the basic idea of this trick. In particular, you should be able to solve the following problem (it's almost identical to the first problem in zscoder's tutorial):

StatusSourceProblem NameDifficultyTags
CSESEasy
Show TagsSlope Trick

It's ok if you found the explanations confusing; the example below should help clarify.

## Buy Low Sell High

Focus Problem – read through this problem before continuing!

### Slow Solution

Let $dp[i][j]$ denote the maximum amount of money you can have on day $i$ if you have exactly $j$ shares of stock on that day. The final answer will be $dp[N]$. This solution runs in $\mathcal{O}(N^2)$ time.

Slow Code

If we run this on the first sample case, then we get the following table:

Input:

9
10 5 4 7 9 12 6 2 10

Output:

dp = {  0}
dp = {  0, -10}
dp = {  0,  -5, -15}
dp = {  0,  -4,  -9, -19}
dp = {  3,  -2,  -9, -16, -26}
dp = {  7,   0,  -7, -16, -25, -35}
dp = { 12,   5,  -4, -13, -23, -35, -47}
dp = { 12,   6,  -1, -10, -19, -29, -41, -53}
dp = { 12,  10,   4,  -3, -12, -21, -31, -43, -55}
dp = { 20,  14,   7,  -2, -11, -21, -31, -41, -53, -65}

However, the DP values look quite special! Specifically, let

$dif[i][j]=dp[i][j]-dp[i][j+1]\ge 0.$

Then $dif[i][j]\le dif[i][j+1]$ for all $j\ge 0$. In other words, $dp[i][j]$ as a function of $j$ is concave down.

Explanation

My Code

### Extension

Stock Trading (USACO Camp): What if your amount of shares can go negative, but you can never have more than $L$ shares or less than $-L$?

## Potatoes & Fertilizers

Focus Problem – read through this problem before continuing!

### Simplifying the Problem

Instead of saying that moving fertilizer from segment $i$ to segment $j$ costs $|i-j|$, we'll say that it costs $1$ to move fertilizer from a segment to an adjacent segment.

Let the values of $a_1,a_2,\ldots,a_N$ after all the transfers be $a_1',a_2',\ldots,a_N'$. If we know this final sequence, how much did the transfers cost (in the best case scenario)? It turns out that this is just

$C=\sum_{i=1}^{N-1}\left|\sum_{j=1}^i(a_j-a_j')\right|.$

We can show that this is a lower bound and that it's attainable. The term $D=\sum_{j=1}^i(a_j-a_j')$ denotes the number of units of fertilizer that move from segment $i$ to segment $i+1$. Namely, if $D$ is positive then $D$ units of fertilizer moved from segment $i$ to segment $i+1$; otherwise, $-D$ units of fertilizer moved in the opposite direction. Note that it is never optimal to have fertilizer moving in both directions.

Let $dif_i=a_i-b_i$ and define $d_j=\sum_{i=1}^jdif_i$ for each $0\le j\le N$. Similarly, define $dif_i'=a_i'-b_i$ and $d_j'=\sum_{i=1}^jdif_i'$. Since we want $dif_i'\ge 0$ for all $i$, we should have $d_0=d_0'\le d_1'\le \cdots\le d_N'=d_N.$ Conversely, every sequence $(d_0',d_1',\ldots,d_N')$ that satisfies this property corresponds to a valid way to assign values of $(a_1',a_2',\ldots,a_N')$.

Now you can verify that $C=\sum_{i=1}^{N-1}|d_i-d_i'|$. This makes sense since moving one unit of fertilizer one position is equivalent to changing one of the $d_i$ by one (although $d_0,d_N$ always remain the same).

### Slow Solution

For each $0\le i\le N$ and $0\le j\le d_N$, let $dp[i][j]$ be the minimum cost to determine $d_0',d_1',\ldots,d_i'$ such that $d_i'\le j$. Note that by definition, $dp[i][j]\ge dp[i][j+1]$. We can easily calculate these values in $\mathcal{O}(N\cdot d_N)$ time.

Explanation

My Code

## USACO Landscaping

Focus Problem – read through this problem before continuing!

This looks similar to the previous task (we're moving dirt instead of fertilizer), so it's not too hard to guess that slope trick is applicable.

### Slow Solution

Let $dp[i][j]$ equal the number of ways to move dirt around the first $i$ flowerbeds such that the first $i-1$ flowerbeds all have the correct amount of dirt while the $i$-th flowerbed has $j$ extra units of dirt (or lacks $-j$ units of dirt if $j$ is negative). The answer will be $dp[N]$.

Explanation

My Solution

### Extension

We can solve this problem when $\sum A_i+\sum B_i$ is not so small by maintaining a map from $j$ to $dif[j+1]-dif[j]$ for all $j$ such that the latter quantity is nonzero. Then the operation "add $Z\cdot |j|$ to $DP[j]$ for all $j$" corresponds to a point update in the map (advance() in the code below).

Code (from Alex Wei)

## Problems

Although we haven't provided any examples of this, some of the problems below will require you to merge two slope containers (usually priority queues).

StatusSourceProblem NameDifficultyTags
CFNormal
Show TagsSlope Trick
CCNormal
Show TagsSlope Trick
CEOINormal
Show TagsSlope Trick
NOI.sgHard
Show TagsSlope Trick
CFHard
Show TagsSlope Trick
CFHard
Show TagsSlope Trick
APIOHard
Show TagsSlope Trick, Small to Large
CFVery Hard
Show TagsSlope Trick
ICPC WFVery Hard
Show TagsSlope Trick, Small to Large

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