(Analysis by Eric Wei)

First, note that it is sometimes possible to change a letter to a different one in faster time than what is originally given. To deal with this, we run Floyd-Warshall on the letters to determine the actual shortest amount of time to change each letter to any other letter.

We can now store the fastest time the letter in each index $i$ can be changed to a different letter $j$, which we represent in a cost matrix, $cst[i][j]$. After this, we can now utilize prefix sums to speed up future computation. We store this in $ps[i][j]$, representing the sum of all $cst[k][j]$ for $k \le i$.

Finally, we can use DP. Let $dp[i][j]$ represent the smallest amount of time needed so that the first $i$ letters create a “valid combo” and the last letter is $j$ (so the last $K$ letters are all $j$). We also create an array, $dpmin[i]$, which represents the minimum value of all $dp[i][j]$ across all possible $j$. The transition is as follows:

dp[i][j] = min(ps[i][j]-ps[i-K][j]+dpmin[i-K],dp[i-1][j]+cst[i][j]);

There are two possibilities we must consider: either the rightmost streak is exactly $K$ letters long, or it is more than that, corresponding to these two possibilities. Calculating $dpmin$ is easy from this, and the final answer is simply $dpmin[N]$. The runtime is $O(M^3+N\cdot M)$. My code follows:

#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for (int i=(a); i<=(signed)(b); i++)
#define F0R(i, a) for (int i=0; i<(signed)(a); i++)
#define MN 100005
#define MA 26
int n, m, k;
string s;
int d[MA][MA];
int cst[MN][MA];
int ps[MN][MA];
int dp[MN][MA];
int mn[MN];
int main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    freopen("cowmbat.in", "r", stdin);
    freopen("cowmbat.out", "w", stdout);
    cin >> n >> m >> k >> s;
    F0R(i, m) F0R(j, m) cin >> d[i][j];
    F0R(c, m) F0R(a, m) F0R(b, m)
        d[a][b] = min(d[a][b], d[a][c]+d[c][b]);
    FOR(i, 1, n){
        F0R(j, m){
            cst[i][j] = d[s[i-1]-'a'][j];
            ps[i][j] = ps[i-1][j]+cst[i][j];
            //cout << i << " " << j << " " << cst[i][j] << "\n";
    memset(dp, 0x3f, sizeof dp);
    memset(mn, 0x3f, sizeof mn);
    mn[0] = 0;
    FOR(i, 1, n){
        F0R(j, m){
            dp[i][j] = min(dp[i][j], dp[i-1][j]+cst[i][j]);
            if (i >= k) dp[i][j] = min(dp[i][j], ps[i][j]-ps[i-k][j]+mn[i-k]);
            //cout << "dp " << i << " " << j << " " << dp[i][j] << "\n";
            mn[i] = min(mn[i], dp[i][j]);

    cout << mn[n] << "\n";