题面
题解
很像最长不下降子序列对吧(废话)
设$up[i]$和$down[i]$分别表示$i$最大最小能取多少
注意到:
$$ f[i] = max_jleft{f[j] ight} + 1 \ a[j] leq down[i],; up[j] leq a[i],; j leq i $$
三位偏序!!!
$CDQ$分治走起
代码
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
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;
}
using std::max;
const int maxn(100010);
int n, m, a[maxn], up[maxn], down[maxn], c[maxn], f[maxn], Max;
inline bool cmp_1(int x, int y) { return a[x] < a[y]; }
inline bool cmp_2(int x, int y) { return down[x] < down[y]; }
void clean(int x) { while(x <= Max) c[x] = 0, x += x & -x; }
void add(int x, int v) { while(x <= Max) c[x] = max(c[x], v), x += x & -x; }
int query(int x)
{
int ans = 0;
while(x) ans = max(ans, c[x]), x -= x & -x;
return ans;
}
void Div(int l, int r)
{
static int id[maxn];
if(l == r) return (void)(f[l] = max(f[l], 1));
int mid = (l + r) >> 1; Div(l, mid);
for(RG int i = l; i <= r; i++) id[i] = i;
std::sort(id + l, id + mid + 1, cmp_1);
std::sort(id + mid + 1, id + r + 1, cmp_2);
int j = l;
for(RG int i = mid + 1; i <= r; i++)
{
while(j <= mid && a[id[j]] <= down[id[i]]) add(up[id[j]], f[id[j]]), j++;
f[id[i]] = max(f[id[i]], query(a[id[i]]) + 1);
}
for(RG int i = l; i < j; i++) clean(up[id[i]]);
Div(mid + 1, r);
}
int main()
{
n = read(), m = read();
for(RG int i = 1; i <= n; i++) up[i] = down[i] = a[i] = read();
for(RG int i = 1, x, y; i <= m; i++)
x = read(), y = read(),
up[x] = max(up[x], y),
down[x] = std::min(down[x], y);
Max = *std::max_element(up + 1, up + n + 1);
Div(1, n); printf("%d
", *std::max_element(f + 1, f + n + 1));
return 0;
}