Submission #759184

#TimeUsernameProblemLanguageResultExecution timeMemory
759184BlagojMecho (IOI09_mecho)C++14
100 / 100
144 ms7268 KiB
#include <bits/stdc++.h>
 
using namespace std;
 
#define endl '\n'
#define ll long long
#define all(x) x.begin(), x.end()
 
ll n, s, times[810][810];
vector<string> v(810);
pair<ll, ll> st, en;
 
const ll dx[4] = {0, 0, -1, 1};
const ll dy[4] = {-1, 1, 0, 0};
 
inline bool valid(ll x, ll y) {
    return (x >= 0 && x < n && y >= 0 && y < n && v[x][y] != 'T');
}
 
bool check(ll m) {
    if (!(m < times[st.first][st.second])) return 0;
    queue<pair<ll, ll>> q;
    q.push(st);
    bool vis[n + 2][n + 2];
    memset(vis, 0, sizeof(vis));
    vis[st.first][st.second] = 1;
    ll moves = 0;
    do {
        ll sz = q.size();
        while (sz--) {
            ll x = q.front().first, y = q.front().second;
            q.pop();
            for (ll d = 0; d < 4; d++) {
                ll nx = x + dx[d], ny = y + dy[d];
                if (valid(nx, ny) && ((m + (moves + 1 == s)) < times[nx][ny] || ((m + (moves + 1 == s)) == times[nx][ny] && make_pair(nx, ny) == en)) && !vis[nx][ny]) {
                    vis[nx][ny] = 1;
                    q.push({nx, ny});
                    if (make_pair(nx, ny) == en) return 1;
                }
            }
        }
        moves++;
        if (moves >= s) {
            moves = 0;
            m++;
        }
    }
    while (q.size() > 0);
    return 0;
}
 
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> s;
    queue<pair<ll, ll>> h;
 
    for (ll i = 0; i < n; i++) {
        cin >> v[i];
        for (ll j = 0; j < n; j++) {
            times[i][j] = n * n + 1000;
            if (v[i][j] == 'M') st = {i, j};
            if (v[i][j] == 'H') {
                h.push({i, j});
                times[i][j] = 0;
            }
            if (v[i][j] == 'D') en = {i, j};
        }
    }
 
    ll mx = 0;
    while (h.size() > 0) {
        ll x = h.front().first, y = h.front().second;
        h.pop();
        for (ll d = 0; d < 4; d++) {
            ll nx = x + dx[d], ny = y + dy[d];
            if (valid(nx, ny) && times[x][y] + 1 < times[nx][ny] && v[nx][ny] != 'D') {
                times[nx][ny] = times[x][y] + 1;
                mx = max(mx, times[nx][ny]);
                h.push({nx, ny});
            }
        }
    }

    ll l = -1, r = mx + 1;
    while (l + 1 < r) {
		ll m = (l + r) / 2;
		if (check(m)) l = m;
		else r = m;
	}
 
    cout << l;
}
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