Loj 2331 某位歌姬的故事

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dp 计数.

首先可以对坐标离散化,记录一下离散化后的每个点代表原来数列中的几个点,方便后续计算方案数.

考虑 $\max(l,r) = h$ 的限制,等价于对于 $l\le i\le r$ 的 $i$ ,都有 $a_i\le h$ ,并且 $[l,r]$ 内至少有一个 $a_i=h$ .

可以对每个点计算出它能取到的上限 $up_i$ ,这个值其实就是所有覆盖它的区间中最小的 $h$ .

可以发现, $[l,r]$ 内至少有一个 $a_i=h$ 的限制,因为 $[l,r]$ 内一定有 $up_i\le h$ ,所以只能由 $up_i=h$ 的 $a_i$ 来满足.

对于所有 $h$ 相同的限制,我们将它们以及 $up_i=h$ 的点一起拿出来单独计算对方案数的贡献.

这需要满足每个区间中,至少有一个 $a_i$ 取到了上限 $a_i=h$ .

如果一个区间完全覆盖了另一个区间,可以直接把大区间的限制去掉,那么剩下的区间一定都是没有包含关系的.

我们把这些区间按照左端点从小到大排序.

设 $dp(i,j)$ 表示确定了前 $i$ 个点是否取到上限,前 $j$ 个区间已经满足限制,但第 $j+1$ 个区间不满足限制的方案数.

最后还要乘上没有被任何区间限制的点的贡献.

时间复杂度 $O(Tm^2)$ .

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#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
int read()
{
	int out = 0, sgn = 1;
	char jp = getchar();
	while (jp != '-' && (jp < '0' || jp > '9'))
		jp = getchar();
	if (jp == '-')
		sgn = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * sgn;
}
const int P = 998244353;
int add(int a, int b)
{
	return a + b >= P ? a + b - P : a + b;
}
void inc(int &a, int b)
{
	a = add(a, b);
}
int mul(int a, int b)
{
	return 1LL * a * b % P;
}
int fpow(int a, int b)
{
	int res = 1;
	while (b)
	{
		if (b & 1)
			res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}
const int N = 2e3 + 10;
int n, m, V, l[N], r[N], h[N], val[N], len, cnt[N], up[N];
struct Interval
{
	int L, R, mx;
	bool operator < (const Interval &rhs) const
	{
		return L ^ rhs.L ? L < rhs.L : R > rhs.R;
	}
} p[N], tmp[N];
struct node
{
	int pos, cnt; 
} q[N];
int dp[N][N];
bool in(node A, Interval B)
{
	return B.L <= A.pos && A.pos <= B.R;
}
int calc(int _n, int _m)
{
	if (!_n)
		return 0;
	sort(tmp + 1, tmp + 1 + _m);
	int sl = 0, minr = n + 1;
	for (int i = _m; i >= 1; --i)
	{
		if (tmp[i].R >= minr)
			continue;
		p[++sl] = tmp[i];
		minr = tmp[i].R;
	}
	_m = sl;
	reverse(p + 1, p + 1 + _m);
	dp[0][0] = 1;
	int mx = p[1].mx;
	for (int i = 0; i < _n; ++i)
	{
		for (int j = 0; j <= _m; ++j)
			if (dp[i][j])
			{
				int c = fpow(mx - 1, q[i + 1].cnt);
				inc(dp[i + 1][j], mul(dp[i][j], c));
				c = add(fpow(mx, q[i + 1].cnt), P - c);
				int pp = j;
				while (pp + 1 <= _m && in(q[i + 1], p[pp + 1]))
					++pp;
				inc(dp[i + 1][pp], mul(dp[i][j], c));
			}
	}
	int res = dp[_n][_m];
	for (int i = 0; i <= _n; ++i)
		for (int j = 0; j <= _m; ++j)
			dp[i][j] = 0;
	return res;
}
pair<int, int> t[N];
void solve()
{
	len = 0;
	n = read(), m = read(), V = read();
	if (m == 0)
	{
		printf("%d\n", fpow(V, n));
		return;
	}
	int ans = 1;
	for (int i = 1; i <= m; ++i)
	{
		l[i] = read(), r[i] = read(), h[i] = read();
		val[++len] = l[i], val[++len] = r[i];
	}
	for (int i = 1; i <= m; ++i)
		for (int j = i + 1; j <= m; ++j)
			if (h[i] > h[j])
			{
				swap(l[i], l[j]);
				swap(r[i], r[j]);
				swap(h[i], h[j]);
			}
	sort(val + 1, val + 1 + len);
	len = unique(val + 1, val + 1 + len) - val - 1;
	ans = mul(ans, fpow(V, val[1] - 1 + n - val[len]));
	int tl = len;
	for (int i = 1; i <= len; ++i)
	{
		cnt[i] = 1;
		if (i < len && val[i] + 1 != val[i + 1])
		{
			++tl;
			val[tl] = val[i] + 1;
			cnt[tl] = val[i + 1] - val[i] - 1;
		}
	}
	len = tl;
	for (int i = 1; i <= len; ++i)
		for (int j = i + 1; j <= len; ++j)
			if (val[i] > val[j])
			{
				swap(val[i], val[j]);
				swap(cnt[i], cnt[j]);
			}
	for (int i = 1; i <= len; ++i)
		up[i] = V;
	for (int i = 1; i <= m; ++i)
	{
		l[i] = lower_bound(val + 1, val + 1 + len, l[i]) - val;
		r[i] = lower_bound(val + 1, val + 1 + len, r[i]) - val;
		for (int j = l[i]; j <= r[i]; ++j)
			up[j] = min(up[j], h[i]);
	}
	for (int i = 1; i <= len; ++i)
		t[i] = make_pair(up[i], i);
	sort(t + 1, t + 1 + len);
	int i = 1, j = 1, _n, _m;
	while (ans && j <= m)
	{
		_m = 0;
		int pp = j, tv = h[j];
		while (pp <= m && h[pp] == tv)
			tmp[++_m] = (Interval){l[pp], r[pp], h[pp]}, ++pp;
		j = pp;
		_n = 0;
		pp = i;
		while (pp <= len && t[pp].first == tv)
			q[++_n] = (node){t[pp].second, cnt[t[pp].second]}, ++pp;
		i = pp;
		ans = mul(ans, calc(_n, _m));			
	}
	while (i <= len)
		ans = mul(ans, fpow(V, cnt[t[i++].second]));
	printf("%d\n", ans);
}
int main()
{
	int T = read();
	while (T--)
		solve();
	return 0;
}