linear DP AcWing 896. Longest ascending subsequence II

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AcWing 896. Longest ascending subsequence II

Algorithm tags

greedy

Ideas

For length is 1 Suborder of , Can be stored a[1], It can also be stored a[2], however a[2] Is better than a[1], Because if the subsequent numbers can be compared with a[1] Constitute ascending subsequence , Then it must be possible to a[2] Constitute ascending subsequence , Therefore, the storage length is i The minimum value at the end of the ascending sequence , Using the method of counter evidence, we can get that the ending value rises with the length of the longest rising sequence .

simulation

Time complexity

For ordered arrays , Find the maximum value less than a certain number , You can use dichotomy , Search efficiency log(n), n Queries , The time complexity is $n\ast log(n)$

Code

#include<bits/stdc++.h>
#define int long long
#define rep(i, a, b) for(int i=a;i<b;++i)
#define Rep(i, a, b) for(int i=a;i>b;--i)
using namespace std;
const int N = 100005, INF = 0x3f3f3f3f;
int a[N], q[N];
inline int read(){
   int s=0,w=1;
   char ch=getchar();
   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
   return s*w;
}
void put(int x) {
    if(x<0) putchar('-'),x=-x;
    if(x>=10) put(x/10);
    putchar(x%10^48);
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n=read();
    rep(i, 0, n){
        a[i]=read();
    }
    int len=0;
    rep(i, 0, n){
        int l=0, r=len;
        while(l<r){
            int mid=l+r+1>>1;
            if(q[mid]<a[i]){
                l=mid;
            }else{
                r=mid-1;
            }
        }
        len=max(len, r+1);
        q[r+1]=a[i];
    }
    printf("%lld", len);
    return 0;
}

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