Submission #1089785

#	Time	Username	Problem	Language	Result	Execution time	Memory
1089785		_callmelucian	Maze (JOI23_ho_t3)	C++14	0 / 100	235 ms	4188 KiB

Main

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

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pl;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tt;

#define all(a) a.begin(), a.end()
#define filter(a) a.erase(unique(all(a)), a.end())

int dist[1010][1010];
bool vis[1010][1010], block[1010][1010];

const int dr[4] = {-1, 0, 0, 1};
const int dc[4] = {0, -1, 1, 0};

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n, m, k; cin >> n >> m >> k;
    int stx, sty, enx, eny; cin >> stx >> sty >> enx >> eny;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            char c; cin >> c;
            if (c == '#') block[i][j] = 1;
            dist[i][j] = INT_MAX;
        }
    }

    function<bool(int,int)> ok = [&] (int i, int j) {
        if (i < 1 || j < 1) return 0;
        if (i > n || j > m) return 0;
        return 1;
    };

    function<int(int,int,int,int)> distance = [&] (int i, int j, int u, int v) {
        return max(abs(i - u), abs(j - v));
    };

    dist[stx][sty] = 0;
    priority_queue<tt> pq;
    pq.emplace(0, stx, sty);

    while (pq.size()) {
        int d, a, b; tie(d, a, b) = pq.top(); pq.pop();
        if (vis[a][b]) continue;
        vis[a][b] = 1;

        //cout << "visting " << a << " " << b << " " << dist[a][b] << "\n";

        function<void(int,int,int)> transition = [&] (int i, int j, int w) {
            if (dist[a][b] + w < dist[i][j]) {
                dist[i][j] = dist[a][b] + w;
                pq.emplace(-dist[i][j], i, j);
            }
        };

        if (block[a][b]) {
            if (distance(a, b, enx, eny) <= k)
                return cout << dist[a][b], 0;

            int xL = max(1, a - k), xR = min(n, a + k);
            for (int j = max(1, b - k + 1); j <= min(m, b + k - 1); j++)
                transition(xL, j, block[xL][j]), transition(xR, j, block[xR][j]);

            int yL = max(1, b - k), yR = min(m, b + k);
            for (int i = max(1, a - k + 1); i <= min(n, a + k - 1); i++)
                transition(i, yL, block[i][yL]), transition(i, yR, block[i][yR]);
        }
        else {
            if (distance(a, b, enx, eny) == 0)
                return cout << dist[a][b], 0;
            for (int k = 0; k < 4; k++)
                if (ok(a + dr[k], b + dc[k])) transition(a + dr[k], b + dc[k], block[a + dr[k]][b + dc[k]]);
        }
    }
    cout << -1;

    return 0;
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 19
• Subtask 3: 0 / 16
• Subtask 4: 0 / 19
• Subtask 5: 0 / 5
• Subtask 6: 0 / 19
• Subtask 7: 0 / 8
• Subtask 8: 0 / 6