POI2011LIZ-Lollipop

POI #Year2011 #构造 #妙妙题

假设能取到 \(x\)，那么 \(\forall y\) , \(x,y\) 奇偶性相同，\(x>y\) ，\(y\) 一定可以是 \(x\) 的一个子区间，处理奇数和偶数的最大值，离线，从大到小做

// Author: xiaruize
const int N = 1e6 + 10;

int n, m;
int a[N], s[N];
pii st, en;
pii qry[N];
pii res[N];
struct mx
{
	int l, r, val;
} od, ev;

void solve()
{
	cin >> n >> m;
	rep(i, 1, n)
	{
		char c;
		cin >> c;
		if (c == 'T')
			a[i] = 2;
		else
			a[i] = 1;
		s[i] = a[i];
		a[i] += a[i - 1];
		if (!st.first && (a[i] & 1))
			st.first = i;
		if (a[i] & 1)
			en.first = i;
		else
			en.second = i;
	}
	rep(i, 1, m)
	{
		cin >> qry[i].first;
		qry[i].second = i;
	}
	sort(qry + 1, qry + m + 1);
	reverse(qry + 1, qry + m + 1);
	if (a[en.second] - a[st.first] > a[en.first] - a[st.second])
		od = {st.first + 1, en.second, a[en.second] - a[st.first]};
	else
		od = {st.second + 1, en.first, a[en.first] - a[st.second]};
	if (a[en.first] - a[st.first] > a[en.second] - a[st.second])
		ev = {st.first + 1, en.first, a[en.first] - a[st.first]};
	else
		ev = {st.second + 1, en.second, a[en.second] - a[st.second]};
	rep(i, 1, m)
	{
		int x = qry[i].first;
		if (x & 1)
		{
			if (x > od.val)
			{
				res[qry[i].second] = {-1, -1};
				continue;
			}
			auto [l, r, val] = od;
			while (val > x)
			{
				if (s[l] == 2)
					l++;
				else if (s[r] == 2)
					r--;
				else
					l++, r--;
				val -= 2;
			}
			res[qry[i].second] = {l, r};
			od = {l, r, val};
		}
		else
		{
			if (x > ev.val)
			{
				res[qry[i].second] = {-1, -1};
				continue;
			}
			auto [l, r, val] = ev;
			while (val > x)
			{
				if (s[l] == 2)
					l++;
				else if (s[r] == 2)
					r--;
				else
					l++, r--;
				val -= 2;
			}
			res[qry[i].second] = {l, r};
			ev = {l, r, val};
		}
	}
	rep(i, 1, m)
	{
		if (res[i].first == -1)
			cout << "NIE" << endl;
		else
			cout << res[i].first << ' ' << res[i].second << endl;
	}
}

#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif

signed main()
{
	// freopen(".in","r",stdin);
	// freopen(".out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int testcase = 1;
	// cin >> testcase;
	while (testcase--)
		solve();
#ifndef ONLINE_JUDGE
	cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
	cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
	return 0;
}