POI #Year2009 #线段树 #Hall定理

考虑实际上是一个二分图匹配问题，那么这个二分图存在匹配当且仅对于 \(L\) 的任何子集右侧的度数和 \(\geq\) 左侧的

然后线段树维护左侧的区间最大和

// Author: xiaruize
const int N = 2e5 + 10;

int n, m, k, d;

struct segment_tree
{
#define ls p << 1
#define rs p << 1 | 1
	struct node
	{
		int mx, lmx, rmx, sum;
	} tr[N << 2];

	void pushup(int p)
	{
		tr[p].lmx = max(tr[ls].lmx, tr[ls].sum + tr[rs].lmx);
		tr[p].rmx = max(tr[rs].rmx, tr[rs].sum + tr[ls].rmx);
		tr[p].mx = max({tr[ls].mx, tr[rs].mx, tr[ls].rmx + tr[rs].lmx});
		tr[p].sum = tr[ls].sum + tr[rs].sum;
	}

	void build(int p, int l, int r)
	{
		if (l == r)
		{
			tr[p] = {-k, -k, -k, -k};
			return;
		}
		int mid = l + r >> 1;
		build(ls, l, mid);
		build(rs, mid + 1, r);
		pushup(p);
	}

	void update(int p, int l, int r, int x, int val)
	{
		if (l == r)
		{
			tr[p].sum += val;
			tr[p].mx += val;
			tr[p].lmx += val;
			tr[p].rmx += val;
			return;
		}
		int mid = l + r >> 1;
		if (x <= mid)
			update(ls, l, mid, x, val);
		else
			update(rs, mid + 1, r, x, val);
		pushup(p);
	}
} seg;

void solve()
{
	cin >> n >> m >> k >> d;
	seg.build(1, 1, n);
	rep(i, 1, m)
	{
		int r, x;
		cin >> r >> x;
		seg.update(1, 1, n, r, x);
		debug(seg.tr[1].mx);
		if (seg.tr[1].mx <= k * d)
			cout << "TAK" << endl;
		else
			cout << "NIE" << 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;
}