이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
 
#define fast ios::sync_with_stdio(0);cin.tie(0);
#define s second
#define f first
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll LOGN = 18; 
const ll INF = 1e15;
const ll MAXN = 2010;
vector<int> occ[2005];
int N, M;
string grid[MAXN];
int emp[MAXN][MAXN], sg[4*MAXN], lazy[4*MAXN];
ll total = 0, ans = 0;
void push(int k, int a, int b) {
	if (lazy[k] != -1) {
		sg[k] = lazy[k];
		if (a != b) {
			lazy[2*k] = lazy[k];
			lazy[2*k+1] = lazy[k];
		}
		lazy[k] = -1;
	}
}
void update(int k, int a, int b, int q_l, int q_r, int new_val) {
	push(k, a, b);
	if (q_r < a || q_l > b)
		return ;
	if (q_l <= a && b <= q_r) {
		lazy[k] = new_val;
		push(k, a, b);
		return ;
	}
	update(2*k, a, (a+b)/2, q_l, q_r, new_val);
	update(2*k+1, (a+b)/2+1, b, q_l, q_r, new_val);
	sg[k] = max(sg[2*k], sg[2*k+1]);
}
ll calc(ll x) {
	return (x * x * x + 3 * x * x + 2 * x) / 6;
}
int get(int k, int a, int b, int pos) {
	push(k, a, b);
	if (b < pos || a > pos)
		return 0;
	if (a == b)
		return sg[k];
	return get(2*k, a, (a+b)/2, pos) + get(2*k+1, (a+b)/2+1, b, pos);
}
int query(int k, int a, int b, int val) {
	push(k, a, b);
	if (a == b) {
		if (sg[k] >= val)
			return a;
		return N+1;
	}
	push(2*k, a, (a+b)/2);
	if (sg[2*k] < val)
		return query(2*k+1, (a+b)/2 + 1, b, val);
	return query(2*k, a, (a+b)/2, val);
}
void split(int pos) {
	int Q = get(1, 0, N, pos);
	int l = query(1, 0, N, Q);
	int r = query(1, 0, N, Q + 1) - 1;
	ans -= calc(r - l);
	ans += calc(pos - l - 1);
	ans += calc(r - pos);
	update(1, 0, N, pos, r, pos);
}
int main() {
	fast
	cin >> N >> M;
	for (int i = 1; i <= N; i++)
		cin >> grid[i];
	for (int j = M; j >= 1; j--) {
		for (int i = 1; i <= N; i++) {
			if (grid[i][j-1] == '.')
				emp[i][j] = emp[i][j+1] + 1;
		}
	}
	memset(sg, 0, sizeof(sg));
	memset(lazy, -1, sizeof(lazy));
	for (int j = 1; j <= M; j++) {
		ans = calc(N);
		for (int i = 1; i <= N; i++) {
			if (emp[i][j] == 0)
				split(i);
			else
				occ[emp[i][j]].push_back(i);
		}
		for (ll q = 1; q <= 2000; q++) {
			total += ans * q;
			while (occ[q].size()) {
				split(occ[q].back());
				occ[q].pop_back();
			}
		}
		update(1, 0, N, 0, N, 0); 
	}
	cout << total << "\n";
}
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