[HEOI2016/TJOI2016]序列

\(\text{Solution}\)

明显有 \(DP\)

\[f_i = f_j + 1(j < i,Mx_j \le a_i,a_j \le Mn_i) \]

然后 \(CDQ\) 分治即可
注意分治顺序

\(\text{Code}\)

#include <cstdio>
#include <iostream>
#include <algorithm>
#define re register
using namespace std;

const int N = 1e5 + 5;
int n, m, a[N], Mx[N], Mn[N], f[N], p[N], len;

inline bool cmp_mx(int x, int y){return Mx[x] < Mx[y];}
inline bool cmp_a(int x, int y){return a[x] < a[y];}

struct BIT{
	int c[N];
	inline int lowbit(int x){return x & (-x);}
	inline void add(int x, int v){for(; x <= len; x += lowbit(x)) c[x] = max(c[x], v);} 
	inline int query(int x){int res = 0; for(; x; x -= lowbit(x)) res = max(res, c[x]); return res;}
	inline void clear(int x){for(; x <= len; x += lowbit(x)) c[x] = 0;}
}T;

void CDQ(int l, int r)
{
	if (l == r) return void(f[l] = max(f[l], 1));
	int mid = l + r >> 1;
	CDQ(l, mid);
	for(re int i = l; i <= r; i++) p[i] = i;
	sort(p + l, p + mid + 1, cmp_mx), sort(p + mid + 1, p + r + 1, cmp_a);
	int j = l - 1;
	for(re int i = mid + 1; i <= r; i++)
	{
		for(; j < mid && Mx[p[j + 1]] <= a[p[i]]; ++j, T.add(a[p[j]], f[p[j]]));
		f[p[i]] = max(f[p[i]], T.query(Mn[p[i]]) + 1);
	}
	for(re int i = l; i <= j; i++) T.clear(a[p[i]]);
	CDQ(mid + 1, r);
}

int main()
{
	scanf("%d%d", &n, &m);
	for(re int i = 1, x; i <= n; i++) scanf("%d", &x), a[i] = Mn[i] = Mx[i] = x, len = max(len, x);
	for(re int i = 1, x, y; i <= m; i++) scanf("%d%d", &x, &y), Mn[x] = min(Mn[x], y), Mx[x] = max(Mx[x], y);
	CDQ(1, n); int ans = 1;
	for(re int i = 1; i <= n; i++) ans = max(ans, f[i]);
	printf("%d\n", ans);
}