BZOJ4361 isn

题面

题解

有难度的计数$dp$

我们先求出所有不降子序列的个数

这个可以用树状数组维护

删除的总方案数为$(n-i)!$种

但是可能我们删到非降之后，我们可能还会删

那么设通过删除操作让子序列变成长度为$i$的方案数为$g[i]$，其中合法的有$f[i]$种

容斥：$f[i] = g[i] - g[i + 1]\times(i + 1)$

就可以啦

代码

#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while(ch != '-' && (!isdigit(ch))) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int maxn(2010), Mod(1e9 + 7), LIM(2000);
int f[maxn][maxn], c[maxn][maxn], fac[maxn], g[maxn], a[maxn], b[maxn], n, m;
inline int Add(int x, int y) { return (x + y) % Mod; }

void add(int a, int b, int c)
{
	for(RG int i = b; i <= LIM; i += i & -i)
		::c[a][i] = Add(::c[a][i], c);
}

int query(int a, int b)
{
	int ans = 0;
	for(RG int i = b; i; i -= i & -i)
		ans = Add(ans, c[a][i]);
	return ans;
}

int main()
{
	n = read(), fac[0] = 1;
	for(RG int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
	for(RG int i = 1; i <= n; i++) a[i] = b[i] = read();
	std::sort(b + 1, b + n + 1);
	m = std::unique(b + 1, b + n + 1) - b - 1;
	for(RG int i = 1; i <= n; i++)
		a[i] = std::lower_bound(b + 1, b + m + 1, a[i]) - b;
	add(0, 1, 1);
	for(RG int i = 1, t; i <= n; i++)
		for(RG int j = i; j; j--)
			t = query(j - 1, a[i]), f[a[i]][j] = Add(f[a[i]][j], t),
			add(j, a[i], t);
	for(RG int i = 1; i <= n; i++)
		for(RG int j = 1; j <= m; j++)
			g[i] = Add(g[i], f[j][i]);
	int ans = 0;
	for(RG int i = 1; i <= n; i++) g[i] = 1ll * fac[n - i] * g[i] % Mod;
	for(RG int i = 1; i <= n; i++)
	{
		if(g[i + 1]) g[i] = (g[i] - 1ll * g[i + 1] * (i + 1) % Mod) % Mod;
		if(g[i]) ans = Add(ans, g[i]);
	}
	if(ans < 0) ans += Mod;
	printf("%d\n", ans);
	return 0;
}