POI2011MET-Meteors

POI #Year2011 #整体二分

整体二分板子，用树状数组维护即可

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

int n, m, t;
vector<int> vec[N];
struct node
{
	int l, r, x;
} s[N];
pii a[N];
struct BIT
{
	int tr[N];

	void add(int x, int v)
	{
		while (x <= m * 2)
		{
			tr[x] += v;
			x += x & -x;
		}
	}

	int sum(int x)
	{
		int res = 0;
		while (x)
		{
			res += tr[x];
			x -= x & -x;
		}
		return res;
	}
} tr;

int res[N];

void solve(int l, int r, int x, int y)
{
	if (x > y)
		return;
	if (l == r)
	{
		debug(l, r, x, y);
		rep(i, x, y)
		{
			res[a[i].second] = l;
			debug(a[i].second);
		}
		return;
	}
	int mid = l + r >> 1;
	rep(i, l, mid)
	{
		tr.add(s[i].l, s[i].x);
		tr.add(s[i].r + 1, -s[i].x);
	}
	vector<pii> lv, rv;
	rep(i, x, y)
	{
		int tmp = 0;
		for (auto v : vec[a[i].second])
		{
			tmp += tr.sum(v) + tr.sum(v + m);
			if (tmp >= a[i].first)
				break;
		}
		debug(l, r, a[i].second, tmp);
		if (tmp >= a[i].first)
			lv.push_back(a[i]);
		else
		{
			a[i].first -= tmp;
			rv.push_back(a[i]);
		}
	}
	rep(i, l, mid)
	{
		tr.add(s[i].l, -s[i].x);
		tr.add(s[i].r + 1, s[i].x);
	}
	int tot = x;
	for (auto v : lv)
		a[tot++] = v;
	for (auto v : rv)
		a[tot++] = v;
	debug(lv, rv);
	solve(l, mid, x, x + lv.size() - 1);
	solve(mid + 1, r, x + lv.size(), y);
}

void solve()
{
	mms(res, -1);
	cin >> n >> m;
	rep(i, 1, m)
	{
		int x;
		cin >> x;
		vec[x].push_back(i);
	}
	rep(i, 1, n)
	{
		cin >> a[i].first;
		a[i].second = i;
	}
	cin >> t;
	rep(i, 1, t)
	{
		cin >> s[i].l >> s[i].r >> s[i].x;
		if (s[i].r < s[i].l)
			s[i].r += m;
	}
	s[++t] = {1, 2 * m, INF};
	solve(1, t, 1, n);
	rep(i, 1, n)
	{
		// debug(res[i]);
		if (res[i] > t - 1 || res[i] == -1)
			cout << "NIE\n";
		else
			cout << res[i] << '\n';
	}
}

#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;
}