题面
解法
显然，本题可以说是变形LIS
设\(f_i\)为以第\(i\)个位置结尾的最长不下降子序列长度
那么，\(f_i=max(f_j+1)\ \ (a_j≤l_i,r_j≤a_i)\)
对于后面的限制条件，显然就是一个二维偏序问题，直接CDQ分治或树套树
然而不知为什么树套树RE了……
代码(树套树50pts)
#include <bits/stdc++.h>
#define N 100010
using namespace std;
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
int lc, rc, mx;
} t[N * 100];
int n, m, tot, siz, l[N], r[N], a[N], f[N], rt[N * 60];
int lowbit(int x) {return x & -x;}
void ins(int &k, int l, int r, int x, int v) {
if (!k) k = ++tot; t[k].mx = max(t[k].mx, v);
if (l == r) return;
int mid = (l + r) >> 1;
if (x <= mid) ins(t[k].lc, l, mid, x, v);
else ins(t[k].rc, mid + 1, r, x, v);
}
void modify(int x, int y, int v) {
for (int i = x; i <= siz; i++)
ins(rt[i], 1, siz, y, v);
}
int query(int k, int l, int r, int x) {
if (!k) return 0;
if (l == r) return t[k].mx;
int mid = (l + r) >> 1;
if (x <= mid) return query(t[k].lc, l, mid, x);
return max(t[t[k].lc].mx, query(t[k].rc, mid + 1, r, x));
}
int ask(int x, int y) {
int ret = 0;
for (int i = x; i; i -= lowbit(i))
ret = max(ret, query(rt[i], 1, siz, y));
return ret;
}
int main() {
read(n), read(m); siz = n;
for (int i = 1; i <= n; i++)
read(a[i]), l[i] = r[i] = a[i], siz = max(siz, a[i]);
for (int i = 1; i <= m; i++) {
int x, y; read(x), read(y);
l[x] = min(l[x], y), r[x] = max(r[x], y);
siz = max(siz, y);
}
int ans = 0;
for (int i = 1; i <= n; i++) {
f[i] = ask(l[i], a[i]) + 1;
modify(a[i], r[i], f[i]);
ans = max(ans, f[i]);
}
cout << ans << "\n";
return 0;
}cdq分治
表示对于这种问题树套树来解决真的辣鸡
cdq分治比较好
时间复杂度：\(O(n\ log^2\ n)\)
代码
#include <bits/stdc++.h>
#define N 200010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Node {
int op, tim, x, y, id;
bool operator < (const Node &a) const {
return x < a.x;
}
} a[N], t[N];
int n, m, mx, v[N], l[N], r[N], ans[N], f[N];
int lowbit(int x) {return x & -x;}
void modify(int x, int v) {
for (int i = x; i <= mx; i += lowbit(i))
if (v == 0) f[i] = v; else chkmax(f[i], v);
}
int query(int x) {
int ret = 0;
for (int i = x; i; i -= lowbit(i))
chkmax(ret, f[i]);
return ret;
}
void cdq(int l, int r) {
if (l == r) return;
int mid = (l + r) >> 1, tx = l, ty = mid + 1;
for (int i = l; i <= r; i++)
if (a[i].tim <= mid) t[tx++] = a[i];
else t[ty++] = a[i];
for (int i = l; i <= r; i++) a[i] = t[i]; cdq(l, mid);
sort(a + l, a + mid + 1);
tx = l, ty = mid + 1;
while (ty <= r) {
while (tx <= mid && a[tx].x <= a[ty].x) {
if (a[tx].op == 1) modify(a[tx].y, ans[a[tx].id]);
tx++;
}
if (a[ty].op == 2) chkmax(ans[a[ty].id], query(a[ty].y) + 1);
ty++;
}
for (int i = l; i < tx; i++) modify(a[i].y, 0);
cdq(mid + 1, r);
}
int main() {
read(n), read(m);
for (int i = 1; i <= n; i++)
read(v[i]), l[i] = r[i] = v[i];
for (int i = 1; i <= m; i++) {
int x, y; read(x), read(y);
chkmin(l[x], y), chkmax(r[x], y);
}
int len = 0; mx = 0;
a[++len] = (Node) {1, len, 1, 1, 0};
for (int i = 1; i <= n; i++) {
a[++len] = (Node) {2, len, v[i], l[i], i};
a[++len] = (Node) {1, len, r[i], v[i], i};
}
for (int i = 1; i <= n; i++) chkmax(mx, r[i]);
sort(a + 1, a + len + 1);
cdq(1, len); int ret = 0;
for (int i = 1; i <= n; i++) chkmax(ret, ans[i]);
cout << ret << "\n";
return 0;
}