【BZOJ4553】[HAOI2016&TJOI2016]序列

【BZOJ4553】[HAOI2016&TJOI2016]序列

题面

bzoj

洛谷

题解

一定要仔细看题啊qwq。。。

我们设$mn[i],mx[i]$表示第$i$个位置上最小出现、最大出现的值。

则选出的序列要满足

$ i<j\\ a[i]\leq mn[j]\\ mx[i]\leq a[j] $

这™不就是个三维偏序吗？

一边$CDQ$一边$dp$就好了

注意分治时注意清空

这题的一个变式

代码

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring> 
#include <cmath> 
#include <algorithm>
using namespace std; 
inline int gi() {
	register int data = 0, w = 1;
	register char ch = 0;
	while (!isdigit(ch) && ch != '-') ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar();
	while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
	return w * data; 
} 
const int MAX_N = 1e5 + 5;
void chkmin(int &x, int y) { if (x > y) x = y; } 
void chkmax(int &x, int y) { if (x < y) x = y; } 
int N, a[MAX_N], mx[MAX_N], mn[MAX_N], f[MAX_N]; 
struct Node { int x, y, z, id; } t[MAX_N];
bool cmp_id (Node a, Node b) { return a.id < b.id; } 
bool cmp_x (Node a, Node b) { return a.x == b.x ? a.y < b.y : a.x < b.x; } 
bool cmp_y (Node a, Node b) { return a.y == b.y ? a.z < b.z : a.y < b.y; }
int c[MAX_N], M; 
inline int lb(int x) { return x & -x; }
void add(int x, int v) { while (x <= M) chkmax(c[x], v), x += lb(x); }
int sum(int x) { int res = 0; while (x > 0) chkmax(res, c[x]), x -= lb(x); return res; }
void Set(int x) { while (x <= M) c[x] = 0, x += lb(x); } 
void Div(int l, int r) { 
	if (l == r) return (void)chkmax(f[t[l].id], 1); 
	int mid = (l + r) >> 1;
	Div(l, mid); 
	sort(&t[l], &t[mid + 1], cmp_x); 
	sort(&t[mid + 1], &t[r + 1], cmp_y); 
	for (int i = mid + 1, j = l; i <= r; i++) { 
		while (t[j].x <= t[i].y && j <= mid) add(t[j].z, f[t[j].id]), ++j; 
		chkmax(f[t[i].id], sum(t[i].x) + 1); 
	}
	for (int i = l; i <= mid; i++) Set(t[i].z);
	sort(&t[l], &t[r + 1], cmp_id); 
	Div(mid + 1, r); 
} 
int main () {
	N = gi(), M = gi(); 
	for (int i = 1; i <= N; i++) a[i] = mn[i] = mx[i] = gi(); 
	while (M--) { 
		int x = gi(), y = gi(); 
		chkmax(mx[x], y), chkmin(mn[x], y); 
	} 
	for (int i = 1; i <= N; i++) t[i] = (Node){a[i], mn[i], mx[i], i}, chkmax(M, mx[i]); 
	Div(1, N); 
	int ans = 0;
	for (int i = 1; i <= N; i++) chkmax(ans, f[i]);
	printf("%d\n", ans); 
	return 0; 
}