(Analysis by Nick Wu)

This problem reads directly as a shortest path problem - the trickiness here is that the length of a path is actually the time that Bessie spends eating grass. Therefore, it may be optimal to take a walk where Bessie eats grass many times, but only for short intervals of time, as opposed to a more direct path where Bessie eats grass fewer times but for longer intervals of time.

The state that we need to keep track of is Bessie's current location and the number of fields Bessie can still travel through before she must eat grass. The transitions out of a state involves moving either horizontally or vertically by one square and possibly spend time eating grass.

Here is Mark Gordon's code.

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

int dr[] = {0, 1, 2, 3, 2, 1, 0, -1, -2, -3, -2, -1, -1, 1, 0, 0};
int dc[] = {3, 2, 1, 0, -1, -2, -3, -2, -1, 0, 1, 2, 0, 0, -1, 1};

int main() {
int N, T;
cin >> N >> T;

vector<vector<int> > A(N, vector<int>(N));
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
cin >> A[i][j];
}
}

vector<vector<int> > D(N, vector<int>(N, 0x7FFFFFFF));
D[0][0] = 0;

priority_queue<pair<int, int> > q;
q.push(make_pair(0, 0));

int result = 0x7FFFFFFF;
while (!q.empty()) {
int d = -q.top().first;
int r = q.top().second / N;
int c = q.top().second % N;
q.pop();
if (d != D[r][c]) {
continue;
}

int dist = abs(N - 1 - r) + abs(N - 1 - c);
if (dist <= 2) {
result = min(result, d + dist * T);
}

for (int i = 0; i < sizeof(dr) / sizeof(int); i++) {
int nr = r + dr[i];
int nc = c + dc[i];
if (nr < 0 || nr >= N || nc < 0 || nc >= N ||
D[nr][nc] < d + A[nr][nc] + 3 * T) {
continue;
}
D[nr][nc] = d + A[nr][nc] + 3 * T;
q.push(make_pair(-D[nr][nc], nr * N + nc));
}
}
cout << result << endl;

return 0;
}