最长不下降子序列nlogn

b[i]表示长度为i的最长不下降子序列的最小末尾元素的值
显然它是单调递增的，满足二分性质，然后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;
}