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答案:
#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;
}
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