Submission #212401

#	Time	Username	Problem	Language	Result	Execution time	Memory
212401		mode149256	Mecho (IOI09_mecho)	C++14	68 / 100	225 ms	6500 KiB

mecho

/*input
7 3
TTTTTTT
TGGGGGT
TGGGGGT
MGGGGGD
TGGGGGT
TGGGGGT
TGHHGGT
*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

using vc = vector<char>;
int N, S;
vector<vc> sk;

vii hiveDis;

int dx[] = {0, 1, 0, -1};
int dy[] = { -1, 0, 1, 0};
pi dxy[] = {
	{dx[0], dy[0]},
	{dx[1], dy[1]},
	{dx[2], dy[2]},
	{dx[3], dy[3]}
};
pi mecho;
pi home;
vpi hives;

bool valid(pi a) {
	return a.x >= 0 and a.y >= 0 and a.x < N and a.y < N;
}

bool gali(int time) {
	vii dis(N, vi(N, MOD));

	dis[mecho.x][mecho.y] = 0;
	queue<pi> q;
	q.push(mecho);

	while (q.size()) {
		pi c = q.front(); q.pop();

		for (int i = 0; i < 4; ++i)
		{
			pi n = c + dxy[i];
			if (valid(n) and sk[n.x][n.y] != 'T') {
				if (sk[n.x][n.y] == 'D')
					return true;

				int newDis = dis[c.x][c.y] + 1;

				if (time + newDis / S < hiveDis[n.x][n.y] and newDis < dis[n.x][n.y]) {
					dis[n.x][n.y] = newDis;
					q.push(n);
				}
			}
		}
	}

	return false;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> N >> S;
	sk = vector<vc>(N, vc(N));
	hiveDis = vii(N, vi(N, MOD));


	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < N; ++j)
		{
			cin >> sk[i][j];
			if (sk[i][j] == 'M') mecho = {i, j};
			else if (sk[i][j] == 'D') home = {i, j};
			else if (sk[i][j] == 'H') hives.emplace_back(i, j);
		}
	}

	queue<pi> q;
	for (auto u : hives) {
		hiveDis[u.x][u.y] = 0;
		q.push(u);
	}

	while (q.size()) {
		pi c = q.front(); q.pop();

		for (int i = 0; i < 4; ++i)
		{
			pi nc = c + dxy[i];
			if (valid(nc) and sk[nc.x][nc.y] != 'T' and nc != home) {
				if (hiveDis[c.x][c.y] + 1 < hiveDis[nc.x][nc.y]) {
					hiveDis[nc.x][nc.y] = hiveDis[c.x][c.y] + 1;
					q.push(nc);
				}
			}
		}
	}

	int l = 0;
	int h = MOD;
	int m;

	while (l < h) {
		m = (l + h + 1) / 2;
		if (gali(m))
			l = m;
		else
			h = m - 1;
	}
	printf("%d\n", l);
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 68 / 100