Maximum Building I

Pista

Fijar la fila inferior del rectángulo. Luego aplicar la solución de Advertisement.

Solución 1

Complejidad temporal: $\mathcal{O}(NM)$

Actualizar el árbol más cercano de cada columna para cada nueva fila inferior. Iterar de izquierda a derecha y usar una pila monótona para calcular el área de cada fila inferior.

C++Python


#include <bits/stdc++.h>
using namespace std;
#define pi pair<int, int>
#define f first
#define s second
 
queue<int> col[1005];
char grid[1005][1005];
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < N; i++) {
		string str;
		cin >> str;
		for (int j = 0; j < M; j++) {
			grid[j][i] = str[j];
			if (grid[j][i] == '*') { col[j].push(i); }
		}
	}
 
	for (int j = 0; j < M; j++) col[j].push(N);
 
	int maxa = 0;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			if (col[j].front() < i) col[j].pop();  // actualiza el árbol más cercano
		}
		stack<pi> sta;
		for (int j = 0; j < M; j++) {
			int start = j;
			while (!sta.empty() && col[j].front() - i < sta.top().s) {
				pi cur = sta.top();
				sta.pop();
				start = cur.f;
				maxa = max(maxa, (j - cur.f) * cur.s);
			}
			sta.push(make_pair(start, col[j].front() - i));
		}
		while (!sta.empty()) {
			pi cur = sta.top();
			sta.pop();
			maxa = max(maxa, (M - cur.f) * cur.s);
		}
	}
 
	cout << maxa << endl;
}


n, m = map(int, input().split())
 
grid = [[""] * n for _ in range(m)]
col = [[] for _ in range(m)]
 
for i in range(n):
	str_row = input().strip()
	for j in range(m):
		grid[j][i] = str_row[j]
		if grid[j][i] == "*":
			col[j].append(i)
 
for j in range(m):
	col[j].append(n)
 
maxa = 0
for i in range(n):
	for j in range(m):
		if col[j] and col[j][0] < i:  # actualiza el árbol más cercano
			del [col[j][0]]
 
	stack = []
	for j in range(m):
		start = j
		while stack and col[j][0] - i < stack[-1][1]:
			cur = stack.pop()
			start = cur[0]
			maxa = max(maxa, (j - cur[0]) * cur[1])
		stack.append((start, col[j][0] - i))
 
	while stack:
		cur = stack.pop()
		maxa = max(maxa, (m - cur[0]) * cur[1])
 
print(maxa)

Solución 2

Complejidad temporal: $\mathcal{O}(NM)$

Igual que la solución 1, pero podemos usar DSU en vez de una pila monótona.

C++


#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
using ld = long double;
using db = double;
using str = string;  // yay python!
 
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
 
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
 
#define tcT template <class T
#define tcTU tcT, class U
// ^ lol this makes everything look weird but I'll try it
tcT > using V = vector<T>;
tcT, size_t SZ > using AR = array<T, SZ>;
tcT > using PR = pair<T, T>;
 
// pairs
#define mp make_pair
#define f first
#define s second
 
// vectors
// oops size(x), rbegin(x), rend(x) need C++17
#define sz(x) int((x).size())
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define rall(x) x.rbegin(), x.rend()
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pb push_back
#define eb emplace_back
#define pf push_front
 
#define lb lower_bound
#define ub upper_bound
tcT > int lwb(V<T> &a, const T &b) { return int(lb(all(a), b) - bg(a)); }
 
// loops
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (int i = (b) - 1; i >= (a); --i)
#define R0F(i, a) ROF(i, 0, a)
#define trav(a, x) for (auto &a : x)
 
const int MOD = 1e9 + 7;  // 998244353;
const int MX = 2e5 + 5;
const ll INF = 1e18;  // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};  // for every grid problem!!
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
 
// bitwise ops
// also see https://gcc.gnu.org/onlinedocs/gcc/Other-Builtins.html
constexpr int pct(int x) { return __builtin_popcount(x); }  // # of bits set
constexpr int bits(int x) {  // assert(x >= 0); // make C++11 compatible until
	                         // USACO updates ...
	return x == 0 ? 0 : 31 - __builtin_clz(x);
}  // floor(log2(x))
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
 
ll cdiv(ll a, ll b) {
	return a / b + ((a ^ b) > 0 && a % b);
}  // divide a by b rounded up
ll fdiv(ll a, ll b) {
	return a / b - ((a ^ b) < 0 && a % b);
}  // divide a by b rounded down
 
tcT > bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }  // set a = min(a,b)
tcT > bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
 
tcTU > T fstTrue(T lo, T hi, U f) {
	hi++;
	assert(lo <= hi);  // assuming f is increasing
	while (lo < hi) {  // find first index such that f is true
		T mid = lo + (hi - lo) / 2;
		f(mid) ? hi = mid : lo = mid + 1;
	}
	return lo;
}
tcTU > T lstTrue(T lo, T hi, U f) {
	lo--;
	assert(lo <= hi);  // assuming f is decreasing
	while (lo < hi) {  // find first index such that f is true
		T mid = lo + (hi - lo + 1) / 2;
		f(mid) ? lo = mid : hi = mid - 1;
	}
	return lo;
}
tcT > void remDup(vector<T> &v) {  // sort and remove duplicates
	sort(all(v));
	v.erase(unique(all(v)), end(v));
}
tcTU > void erase(T &t, const U &u) {  // don't erase
	auto it = t.find(u);
	assert(it != end(t));
	t.erase(it);
}  // element that doesn't exist from (multi)set
 
// INPUT
#define tcTUU tcT, class... U
tcT > void re(complex<T> &c);
tcTU > void re(pair<T, U> &p);
tcT > void re(V<T> &v);
tcT, size_t SZ > void re(AR<T, SZ> &a);
 
tcT > void re(T &x) { cin >> x; }
void re(db &d) {
	str t;
	re(t);
	d = stod(t);
}
void re(ld &d) {
	str t;
	re(t);
	d = stold(t);
}
tcTUU > void re(T &t, U &...u) {
	re(t);
	re(u...);
}
 
tcT > void re(complex<T> &c) {
	T a, b;
	re(a, b);
	c = {a, b};
}
tcTU > void re(pair<T, U> &p) { re(p.f, p.s); }
tcT > void re(V<T> &x) { trav(a, x) re(a); }
tcT, size_t SZ > void re(AR<T, SZ> &x) { trav(a, x) re(a); }
tcT > void rv(int n, V<T> &x) {
	x.rsz(n);
	re(x);
}
 
// TO_STRING
#define ts to_string
str ts(char c) { return str(1, c); }
str ts(const char *s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) {
#ifdef LOCAL
	return b ? "true" : "false";
#else
	return ts((int)b);
#endif
}
tcT > str ts(complex<T> c) {
	stringstream ss;
	ss << c;
	return ss.str();
}
str ts(V<bool> v) {
	str res = "{";
	F0R(i, sz(v)) res += char('0' + v[i]);
	res += "}";
	return res;
}
template <size_t SZ> str ts(bitset<SZ> b) {
	str res = "";
	F0R(i, SZ) res += char('0' + b[i]);
	return res;
}
tcTU > str ts(pair<T, U> p);
tcT > str ts(T v) {  // containers with begin(), end()
#ifdef LOCAL
	bool fst = 1;
	str res = "{";
	for (const auto &x : v) {
		if (!fst) res += ", ";
		fst = 0;
		res += ts(x);
	}
	res += "}";
	return res;
#else
	bool fst = 1;
	str res = "";
	for (const auto &x : v) {
		if (!fst) res += " ";
		fst = 0;
		res += ts(x);
	}
	return res;
 
#endif
}
tcTU > str ts(pair<T, U> p) {
#ifdef LOCAL
	return "(" + ts(p.f) + ", " + ts(p.s) + ")";
#else
	return ts(p.f) + " " + ts(p.s);
#endif
}
 
// OUTPUT
tcT > void pr(T x) { cout << ts(x); }
tcTUU > void pr(const T &t, const U &...u) {
	pr(t);
	pr(u...);
}
void ps() { pr("\n"); }  // print w/ spaces
tcTUU > void ps(const T &t, const U &...u) {
	pr(t);
	if (sizeof...(u)) pr(" ");
	ps(u...);
}
 
// DEBUG
void DBG() { cerr << "]" << endl; }
tcTUU > void DBG(const T &t, const U &...u) {
	cerr << ts(t);
	if (sizeof...(u)) cerr << ", ";
	DBG(u...);
}
#ifdef LOCAL  // compile with -DLOCAL, chk -> fake assert
#define dbg(...)                                                                       \
	cerr << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#define chk(...)                                                                       \
	if (!(__VA_ARGS__))                                                                \
		cerr << "Line(" << __LINE__ << ") -> function(" << __FUNCTION__                \
		     << ") -> CHK FAILED: (" << #__VA_ARGS__ << ")" << "\n",                   \
		    exit(0);
#else
#define dbg(...) 0
#define chk(...) 0
#endif
 
void setPrec() { cout << fixed << setprecision(15); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
// FILE I/O
void setIn(str s) { freopen(s.c_str(), "r", stdin); }
void setOut(str s) { freopen(s.c_str(), "w", stdout); }
void setIO(str s = "") {
	unsyncIO();
	setPrec();
	// cin.exceptions(cin.failbit);
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) setIn(s + ".in"), setOut(s + ".out");  // for USACO
}
 
template <int SZ> struct DSU {
	int par[SZ], sz[SZ];
	DSU() { F0R(i, SZ) par[i] = -1, sz[i] = 1; }
 
	int get(int x) {  // compresión de caminos
		if (par[x] != x) par[x] = get(par[x]);
		return par[x];
	}
 
	bool unite(int x, int y) {  // unión por rango
		x = get(x), y = get(y);
		if (x == y) return 0;
		if (sz[x] < sz[y]) swap(x, y);
		sz[x] += sz[y], par[y] = x;
		return 1;
	}
};
 
int n, m, cur[1000];
char g[1000][1000];
int ans = 0;
 
void solve(int x) {
	vi nex[m + 1];
	F0R(i, n) nex[cur[i] - x].pb(i);
	DSU<1000> D;
	R0F(i, m + 1) for (int a : nex[i]) {
		D.par[a] = a;
		if (a > 0 && D.par[a - 1] != -1) D.unite(a, a - 1);
		if (a < n - 1 && D.par[a + 1] != -1) D.unite(a, a + 1);
		ans = max(ans, i * D.sz[D.get(a)]);
	}
}
 
int solve() {
	F0R(i, n) cur[i] = m;
	R0F(j, m) {
		F0R(i, n) if (g[i][j] == '*') cur[i] = j;
		solve(j);
	}
	return ans;
}
 
int main() {
	setIO();
	cin >> n >> m;
	F0R(i, n) F0R(j, m) cin >> g[i][j];
	cout << solve();
}