(Analysis by Benjamin Qi, Richard Qi)

Call the tiles which are stuck in place "fixed values", and call all other values "unfixed values". First, append large fixed values (say, $10^6$) to either end of the array, so that every unfixed value is between two fixed values. At the end, we then subtract $2 \cdot 10^6$ from the answer.

Define a range to be the space between two consecutive fixed tiles $a_l$ and $a_r$.

Instead of the original scenario of a sequence of tiles, we can imagine that we have $K+1$ "ranges", where the $i$th range consists of two values $m_i = \min(a_l, a_r), M_i = \max(a_l, a_r)$, and some size $len_i$, satisfying $\sum_{i=1}^{K+1} len_i = N-K$. Our job is to allocate $N-K$ values to the $K+1$ ranges, where the $i$th range gets a sequence of values $b_{i, 1}, b_{i, 2}, \dots, b_{i, len_i}$, such that the value we want to minimize is $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \max(b_{i, j}, b_{i, j+1})$ where we also define $b_{i, 0} = m_i$, $b_{i, len_i+1} = M_i$ as fixed values.

Claim: For a range $i$, it is optimal for the values to be sorted such that $b_{i, j} \le b_{i, j+1}$ for $j \in [1, len_i-1]$.

Proof: This holds from the $0-1$ principle. If all numbers are $0$ or $1$, then clearly it is optimal for the values to be sorted in the above manner. So, it is optimal to sort even when the numbers are not $0$ or $1$.

More rigorously, the objective function we want to minimize can be written as $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \sum_{v=0}^{\infty} \max(f_v(b_{i, j}), f_v(b_{i, j+1})) = \sum_{v=0}^{\infty} \sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \max(f_v(b_{i, j}), f_v(b_{i, j+1}))$. where $f_v(x) = 1$ if $x > v$, and otherwise $f_v(x) = 0$. Now, for each value of $v$, the inner summation is minimized when the values are sorted. So, it is optimal to sort the values in increasing order.

So, for the $K = 0$ case, there is only one range, and we simply sort the values to compute $\sum_{j=0}^{len_1} \max(b_{1, j}, b_{1, j+1})$.

For $N \le 50$, we can do a $O(\prod_{i=1}^{K+1} len_i) \le O((\frac{N}{K+1})^{K+1})$ state DP. For each of the unfixed values in increasing value order, we decide which range to put it in, and place the value in the first position which is unfilled in the range. The DP state is the number of elements filled for each range so far $(placed_1, placed_2, \dots, placed_{K+1})$, and the value of the state is the sum over all $i$ of $\sum_{j=0}^{placed_i-1} \max(b_{i, j}, b_{i, j+1})$ if $placed_i < len_i$, and $\sum_{j=0}^{len_i} \max(b_{i, j}, b_{i, j+1})$ if $placed_i = len_i$.

Richard's Code for $N \le 50$:

#include <bits/stdc++.h>
using namespace std;

#define sz(x) int((x).size())

int main() {
cin.tie(0)->sync_with_stdio(0);
int N, K;
cin >> N >> K;
multiset<int> multi_vals;
vector<int> inp_a;
for (int i = 1; i <= N; i++) {
int a;
cin >> a;
multi_vals.insert(a);
inp_a.push_back(a);
}

vector<pair<int, int>> xys;
for (int i = 1; i <= K; i++) {
int x, y;
cin >> x;
y = inp_a[x - 1];
xys.push_back(make_pair(x, y));
multi_vals.erase(multi_vals.find(y));
}
xys.push_back(make_pair(0, 1000000));
xys.push_back(make_pair(N + 1, 1000000));
sort(begin(xys), end(xys));

vector<int> vals;
for (auto u : multi_vals) vals.push_back(u);

// xys and vals store the input

vector<pair<pair<int, int>, int>> rangs; //ranges: ((min value, max value), size)

for (int i = 0; i + 1 < sz(xys); i++) {
int rang_size = xys[i + 1].first - xys[i].first - 1;
if (rang_size == 0) {
addon_zerosz += max(xys[i + 1].second, xys[i].second);
} else {
pair<int, int> rang_vals = make_pair(xys[i].second, xys[i + 1].second);
if (rang_vals.first > rang_vals.second) swap(rang_vals.first, rang_vals.second);
rangs.push_back(make_pair(rang_vals, rang_size));
}
}

map<vector<int>, int> dp;
dp[vector<int>(sz(rangs), 0)] = 0;

for(auto v: vals){ //for each value in increasing order
map<vector<int>, int> ndp;
for(const auto& u: dp){
vector<int> st = u.first; //current # of placed values in each range
for(int i = 0; i < sz(rangs); i++){ //place the next value in range i
if(st[i]+1 <= rangs[i].second){
vector<int> new_st = st;
int new_val = u.second;
new_st[i]++;
if(st[i] == 0){
new_val+=max(v, rangs[i].first.first)-v;
}
if(st[i] == rangs[i].second-1){
new_val+=max(v, rangs[i].first.second)-rangs[i].first.second;
}

if(!ndp.count(new_st) || ndp[new_st] >= new_val){
ndp[new_st] = new_val;
}
}
}
}
swap(dp, ndp);
}

assert(sz(dp) == 1);
int ans = dp.begin()->second;

for (auto u : vals) { ans += u; }
for (auto u : rangs) { ans += u.first.second; }
ans -= 2 * 1000000;
cout << ans << "\n";
}


Now, using the sorted property, notice that in the expression $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \max(b_{i, j}, b_{i, j+1})$, we almost always have $\max(b_{i, j}, b_{i, j+1}) = b_{i, j+1}$. The only $j$ where this might not be the case are $j = 0, len_i$. However, we know the value of $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} b_{i, j+1}$ is just the sum of all non-fixed values plus $\sum_{i=1}^{K+1} M_i$.

So, it suffices to minimize $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \max(b_{i, j}, b_{i, j+1}) - b_{i, j+1}$ instead. From the above observation, we know that $\sum_{i=1}^{K+1} \sum_{j=0}^{len_i} \max(b_{i, j}, b_{i, j+1}) - b_{i, j+1} = \sum_{i=1}^{K+1} \max(b_{i, 0}, b_{i, 1}) - b_{i, 1} + \max(b_{i, len_i}, b_{i, len_i+1})-b_{i, len_i + 1} = \sum_{i=1}^{K+1} \max(m_i, b_{i, 1}) - b_{i, 1} + \max(b_{i, len_i}, M_i)-M_i$

For the $K = 1$ subtask, we have $2$ ranges. Notice that $M_1 = M_2 = +10^6,$ so that for $i = 1, 2$, we have $\max(b_{i, len_i}, b_{i, len_i+1})-b_{i, len_i + 1} = M_i-M_i = 0$. Also, for $K = 1$, we have both $m_1 = m_2 = m$ as the only fixed value. So, we want to minimize $\sum_{i=1}^{2} \max(m, b_{i, 1}) - b_{i, 1}$. Let the unfixed values be $c_1, c_2, \dots, c_{N-1}$, in increasing value order. One of the values $b_{i, 1}$ is forced to be the value $c_1$, and because $\max(m, b_{i, 1}) - b_{i, 1}$ is a decreasing function in $b_{i, 1}$, it suffices to maximize the other $b_{i, 1}$. It is easy to prove that the maximum such value is $c_{\max(len_1, len_2)+1}$, where we allocate a prefix of the values $c_1, c_2, \dots, c_{N-1}$ to be in one range, and a suffix of the values to be in the other range.

Now, we go for the full solution. Define the bounds of a range $i$ to be $[b_{i, 1}, b_{i, len_i}]$. Define the sequence of non-fixed values in increasing order as $c_1, c_2, \dots, c_{N-K}$.

Claim: There is an optimal solution where for every two ranges one of the bounds contains the other, or the bounds do not intersect.

Proof: WLOG, let the two ranges be ranges $1$ and $2$. Let the values currently allocated to range $1$ and $2$ be $c_{s_1} \le c_{s_2} \le \dots \le c_{s_{len_1+len_2}}$ for some sequence of indices $s_1 < s_2 \dots < s_{len_1+len_2}$, such that $len_1$ of the values are allocated to range $1$ and $len_2$ of the values are allocated to range $2$.

Now, if indices $s_1, s_{len_1+len_2}$ both belonged to the same range, then one range would contain the other. Thus, we can assume WLOG that range $1$ contains index $s_1$ and range $2$ contains index $s_{len_1+len_2}$. Also, if first $len_1$ values corresponding to indices $s_1, \dots, s_{len_1}$ all belong to range $1$, then the bounds do not intersect, and we are done.

Otherwise, there is a $k$ such that $s_k$ belongs to range $2$ and $s_{k+1}$ belongs to range $1$. Now, if we can show that we can swap the values such that $s_k$ belongs to range $1$ and $s_{k+1}$ belongs to range $2$, then after making as many of these repeated swaps as possible, we are left in the case where the bounds do not intersect, and we are done. This is because if we list out which range each $s_k$ belongs to as a sequence of $1$s and $2$s, each swap decreases the number of inversions by $1$.

Notice that because $s_1$ corresponds to $b_{1, 1}$ and $s_{len_1+len_2}$ corresponds to $b_{2, len_2}$, $s_k$ does not correspond to $b_{2, len_2}$, and $s_{k+1}$ does not correspond to $b_{1, 1}$. So, the effect of making the swap on $s_k, s_{k+1}$ on $b_{1, 1}, b_{1, len_1}, b_{2, 1}, b_{2, len_2}$ can only result in increasing $b_{2, 1}$ and decreasing $b_{1, len_1}$.

Recall that the quantity for range $i$ that we want to minimize is $\max(m_i, b_{i, 1}) - b_{i, 1} + \max(b_{i, len_i}, M_i)-M_i$, which is decreasing when we increase $b_{i, 1}$ and decrease $b_{i, len_i}$. Thus, making the swap does not decrease the summation, as desired.

Now, we can use dynamic programming, on consecutive subsegments of the array $c_1, c_2, \dots, c_{N-K}$ and a subset of ranges ($O(N^22^K)$ states in total). The state stores the minimum sum of differences attainable if for each range in the subset we only allocate values within the contiguous subsequence to it.

Specifically, $dp[L][R][sub]$ to be the minimum cost of assigning non-fixed values $c_L, c_{L+1}, \dots, c_{R}$ to completely fill the ranges determined by the mask $sub$, where do not necessarily use all of the values $c_L, \dots, c_R$.

First, if the number of required values to fill all ranges in $sub$ is more than $R-L+1$, then we set the cost to $+\infty$. Now, there are three types of transitions:

1. If $c_L$ does not belong to any range in $sub$, we transition to state $dp[L+1][R][sub]$. Similarly, if $c_R$ does not belong to any range in $sub$ we transition to state $dp[L][R-1][sub]$.
2. If $c_L$ and $c_R$ belong to the same range $i$, then we know the value $\max(m_i, b_{i, 1}) - b_{i, 1} + \max(b_{i, len_i}, M_i)-M_i$ for range $i$, which we add to the current cost, and then transition to $dp[L+1][R-1][sub-2^i]$.
3. Else, there must be some split point $M$ such that $c_L, \dots, c_M$ belong to some nonempty subset of ranges $sub_1$, and $c_{M+1}, \dots, c_R$ belong to some nonempty subset of ranges $sub_2$, where $sub_1 \cap sub_2 = \emptyset$, and $sub_1 \cup sub_2 = sub$. So, we transition to $dp[L][M][sub_1] + dp[M+1][R][sub_2]$.

Clearly, we must have $M \ge L+size(sub_1)-1$, where $size(sub_1) = \sum_{i \in sub_1} len_i$. Using a similar swapping argument as the proof of the previous claim, it can be shown that it is optimal for $M = L+size(sub_1)-1$ (in other words, all of the values in the range $[L, M]$ exactly correspond to ranges in $sub_1$ and no other ranges.

The answer is $dp[1][N-K][2^{\text{# of ranges}}-1]$, plus the fixed quantity that we subtracted out earlier.

For each $L, R$ and mask $sub$, we iterate over all submasks of $sub$, for a total time complexity of $O(N^2 3^K)$.

Richard's Code (Full Solution):

#include <bits/stdc++.h>
using namespace std;

#define sz(x) int((x).size())

const int MOD = 1e9+7;

int dp[305][305][1<<7];
int size_sum[1<<7];

int main() {
cin.tie(0)->sync_with_stdio(0);
int N, K; cin >> N >> K;
multiset<int> multi_vals;
vector<int> inp_a;
for(int i = 1; i <= N; i++){
int a; cin >> a; multi_vals.insert(a);
inp_a.push_back(a);
}

vector<pair<int, int>> xys;
for(int i = 1; i <= K; i++){
int x, y; cin >> x; y = inp_a[x-1];
xys.push_back(make_pair(x, y));
multi_vals.erase(multi_vals.find(y));
}
xys.push_back(make_pair(0, 1000000));
xys.push_back(make_pair(N+1, 1000000));
sort(begin(xys), end(xys));

vector<int> vals;
for(auto u: multi_vals) vals.push_back(u);

//xys and vals store the input

vector<pair<pair<int, int>, int>> rangs;
//((min, max), size)
for(int i = 0; i+1 < sz(xys); i++){
int rang_size = xys[i+1].first-xys[i].first-1;
if(rang_size == 0){
}
else{
pair<int, int> rang_vals = make_pair(xys[i].second, xys[i+1].second);
if(rang_vals.first > rang_vals.second) swap(rang_vals.first, rang_vals.second);
rangs.push_back(make_pair(rang_vals, rang_size));
}
}

for(int i = 0; i < (1<<sz(rangs)); i++){
for(int j = 0; j < sz(rangs); j++){
if((i>>j)&1){
size_sum[i]+=rangs[j].second;
}
}
}

for(int i = 0; i < 305; i++) for(int j = 0; j < 305; j++) for(int k = 0; k < (1<<7); k++) dp[i][j][k] = MOD;

for(int L = sz(vals)-1; L >= 0; L--){
for(int R = L; R < sz(vals); R++){
int rang_size = R-L+1;

if(size_sum[mask] <= rang_size-1){ //not use either end
}

//split in half somewhere by some submask
}

for(int k = 0; k < sz(rangs); k++){
//assign L, R to be left and right of range k
if(size_sum[1<<k] == 1 && L < R) continue;

//compute contribution of range k to the sum
int small_val = vals[L];
int big_val = vals[R];
int contrib = max(big_val, rangs[k].first.second)-rangs[k].first.second + max(small_val, rangs[k].first.first)-small_val;
assert(contrib >= 0);

//only thing left
}
else{
assert(L+1 <= R-1);
}
}
}

}
}
}

int ans = dp[0][sz(vals)-1][(1<<sz(rangs))-1];