제출 #328607

#제출 시각아이디문제언어결과실행 시간메모리
328607dolphingarlicNafta (COI15_nafta)C++14
100 / 100
532 ms121192 KiB
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

int n, m;
char grid[2001][2001];
bool visited[2001][2001];
int dx[4]{1, -1, 0, 0}, dy[4]{0, 0, 1, -1};
int bit[2001], diff[2001][2001], dp[2001][2001];

bool inside(int x, int y) {
	return x && y && x <= n && y <= m && !visited[x][y] && grid[x][y] != '.';
}

tuple<int, int, int> get_oil(int x, int y) {
	tuple<int, int, int> ret = {y, y, grid[x][y] - '0'};
	visited[x][y] = true;
	for (int i = 0; i < 4; i++) if (inside(x + dx[i], y + dy[i])) {
		tuple<int, int, int> pot = get_oil(x + dx[i], y + dy[i]);
		get<0>(ret) = min(get<0>(ret), get<0>(pot));
		get<1>(ret) = max(get<1>(ret), get<1>(pot));
		get<2>(ret) += get<2>(pot);
	}
	return ret;
}

void update(int pos, int val) { for (; pos <= m; pos += pos & -pos) bit[pos] += val; }

int query(int pos) {
	int ans = 0;
	for (; pos; pos -= pos & -pos) ans += bit[pos];
	return ans;
}

void solve(int placed, int l = 1, int r = m, int opt_l = 0, int opt_r = m - 1) {
	int mid = (l + r) / 2;
	int opt = -1;
	for (int i = opt_l; i <= min(mid - 1, opt_r); i++) {
		if (dp[placed - 1][i] + diff[mid][i] > dp[placed][mid]) {
			dp[placed][mid] = dp[placed - 1][i] + diff[mid][i];
			opt = i;
		}
	}
	if (l != r) {
		solve(placed, l, mid, opt_l, opt);
		solve(placed, mid + 1, r, opt, opt_r);
	}
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
		cin >> grid[i][j];
	}
	vector<tuple<int, int, int>> oil;
	for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
		if (inside(i, j)) {
			int l, r, val;
			tie(l, r, val) = get_oil(i, j);
			oil.push_back({l, l, val});
			oil.push_back({r + 1, l, -val});
		}
	}
	sort(oil.begin(), oil.end());
	for (int i = 1, ptr = 0; i <= m; i++) {
		while (get<0>(oil[ptr]) <= i) {
			update(get<1>(oil[ptr]), get<2>(oil[ptr]));
			ptr++;
		}
		for (int j = 0; j < i; j++) diff[i][j] = query(i) - query(j);
	}
	for (int i = 1; i <= m; i++) {
		solve(i);
		cout << *max_element(dp[i] + 1, dp[i] + m + 1) << '\n';
	}
	return 0;
}
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