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最長不下降子序列nlogn

b[i]表示長度為i的最長不下降子序列的最小末尾元素的值
顯然它是單調遞增的,滿足二分性質,然後就可以愉快地二分啦.

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<set>
#include<map>
#include<stack>
#include<cstring>
#define inf 2147483647
#define
ls rt<<1 #define rs rt<<1|1 #define lson ls,nl,mid,l,r #define rson rs,mid+1,nr,l,r #define N 100010 #define For(i,a,b) for(int i=a;i<=b;i++) #define p(a) putchar(a) #define g() getchar() using namespace std; int a[N],b[N],n,len; void in(int &x){ int y=1; char c=g();x=0;
while(c<'0'||c>'9'){ if(c=='-')y=-1; c=g(); } while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=g(); } x*=y; } void o(int x){ if(x<0){ p('-'); x=-x; } if(x>9)o(x/10); p(x%10+'0'); } int main(){ in(n); For(i,
1,n)in(a[i]); len=1; b[1]=a[1]; For(i,2,n){ if(a[i]>=b[len])b[++len]=a[i]; else{ int x=upper_bound(b+1, b+len+1, a[i])-b; b[x]=a[i]; } // b[upper_bound(b+1, b+len+1, a[i])-b-1]=a[i]; } o(len); return 0; }