new Problem("CF", "Farm of Monsters", "gym/102538/problem/F", "Hard", false, ["Slope Trick"], ""),
new Problem("CF", "Moving Walkways", "contest/1209/problem/H", "Hard", false, ["Slope Trick"], ""),
new Problem("CF", "April Fools' Problem", "contest/802/problem/O", "Very Hard", false, ["Slope Trick"], "binary search on top of slope trick"),
new Problem("ICPC World Finals", "Conquer the World", "https://icpc.kattis.com/problems/conquertheworld", "Very Hard", false, ["Slope Trick"], "ICPC world finals, 0 solves in contest - \"Potatoes\" on tree!!"),
> - 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. The rest of this module assumes that you are somewhat familiar with at least one of the tutorials mentioned above.
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 $dp[N][0]$. This solution runs in $O(N^2)$ time.
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**.
### Full Solution
<spoiler title="Explanation">
We'll process the shares in order. Suppose that we are currently considering the $i$-th day, where shares are worth $p_i$. We can replace (buy or sell a share) in the statement with (buy, then sell somewhere between 0 and 2 shares).
- If we currently have $j$ shares and overall balance $b$, then after buying, $j$ increases by one and $b$ decreases by $p_i$. So we set $dp[i][j]=dp[i-1][j-1]-p_i$ for all $j$. Note that the differences between every two consecutive elements of $dp[i]$ have not changed.
- If we choose to sell a share, this is equivalent to setting $dp[i][j]=\max(dp[i][j],dp[i][j+1]+p_i)$ for all $j$ at the same time. By the concavity condition, $dp[i][j]=dp[i][j+1]+p_i$ will hold for all $j$ less than a certain threshold while $dp[i][j]$ will remain unchanged for all others. So this is equivalent to inserting $p_i$ into the list of differences while maintaining the condition that the differences are in sorted order.
- So choosing to sell between 0 and 2 shares is represented by adding $p_i$ to the list of differences two times. After that, we should pop the smallest difference in the list because we can't end up with a negative amount of shares.
**Example:** consider the transition from `dp[4]` to `dp[5]`. Note that $p_5=9$.
Start with:
```
dp[4] = { 3, -2, -9, -16, -26}
dif[4] = { 5, 7, 7, 10}
```
After buying one share, $9$ is subtracted from each value and they are shifted one index to the right.
```
dp[5] = { x, -6, -11, -18, -25, -35}
dif[5] = { x, 5, 7, 7, 10}
```
Then we can choose to sell one share at price $9$. The last two DP values remain the same while the others change.
```
dp[5] = { 3, -2, -9, -16, -25, -35}
dif[5] = { 5, 7, 7, 9, 10}
```
Again, we can choose to sell one share at price $9$. The last three DP values remain the same while the others change.
```
dp[5] = { 7, 0, -7, -16, -25, -35}
dif[5] = { 7, 7, 9, 9, 10}
```
(insert diagrams)
</spoiler>
<spoiler title="My Code">
The implementation is quite simple; maintain a priority queue representing $dif[i]$ that allows you to pop the minimum element. After adding $i$ elements, $ans$ stores the current value of $dp[i][i]$. At the end, you add all the differences in $dif[N]$ to go from $dp[N][N]$ to $dp[N][0]$.
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
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).
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 $O(N\cdot d_N)$ time.
Similar to before, this DP is concave up for a fixed $i$! Given a piecewise linear function $f_i(x)$ that takes as input $x$ and outputs $dp[i][x]$, we need to support the following two operations to transform this function into $f_{i+1}$.
- Add $|x-k|$ to the function for some $k$
- Set $f(x)=\min(f(x),f(x-1))$ for all $x$
Again, these can be done with a priority queue. Instead of storing the consecutive differences, we store the points where the slope of the piecewise linear function changes by one.
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 $dp[N][0]$.
As before, it helps to look at the differences $dif[j]=DP[j+1]-DP[j]$ instead. We'll maintain separate deques for $dif$ depending on whether $j < 0$ or $j\ge 0$. We'll call these the left and right deques, respectively.
- The first two operations correspond to repeatedly popping the last element off of the left deque and adding it to the front of the right deque (or vice versa, depending on the direction of the shift).
- The third operation corresponds to subtracting $Z$ from all elements of the left deque and adding $Z$ to all elements of the right deque.
- The last operation corresponds to setting $dif[j]=\max(dif[j],-Y)$ for all $j < 0$