Longest ascending subsequence model AcWing 1014. Mountaineering

Original link

AcWing 1014. Mountaineering

Algorithm tags

DP linear DP Longest ascending subsequence

Ideas

Ask for i Ending ascending sequence , With i The reverse rising at the end is the descending sequence , With i It is a sequence in which the dividing point rises first and then falls , Find the maximum .

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 = 1005, INF = 0x3f3f3f3f;
int f[N], f1[N], a[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, 1, n+1){
	    a[i]=read();
	}
	int ans=0, ans1=0, ans2=0;
	//  With i Ending ascending sequence 
	rep(i, 1, n+1){
	    f[i]=1;
	    rep(j, 1, i){
	        if(a[j]<a[i]){
	            f[i]=max(f[i], f[j]+1);
	        }
	    }
	    ans1=max(ans1, f[i]);
	}
	//  With i The reverse rising at the end is the descending sequence 
	Rep(i, n, 0){
	    f1[i]=1;
	    Rep(j, n, i+1){
	        if(a[j]<a[i]){
	            f1[i]=max(f1[i], f1[j]+1);
	        }
	    }
	    ans2=max(ans2, f1[i]);
	}
	//  Go up first   Post descent sequence 
	rep(i, 1, n+1){
	    ans=max(ans, f[i]+f1[i]-1);
	}
	printf("%lld\n", max({ans, ans1, ans2}));
}

And y The general code is consistent

y Master code

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;

int n;
int h[N];
int f[N], g[N];

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%d", &h[i]);

    for (int i = 0; i < n; i ++ )
    {
        f[i] = 1;
        for (int j = 0; j < i; j ++ )
            if (h[i] > h[j])
                f[i] = max(f[i], f[j] + 1);
    }

    for (int i = n - 1; i >= 0; i -- )
    {
        g[i] = 1;
        for (int j = n - 1; j > i; j -- )
            if (h[i] > h[j])
                g[i] = max(g[i], g[j] + 1);
    }

    int res = 0;
    for (int i = 0; i < n; i ++ ) res = max(res, f[i] + g[i] - 1);

    printf("%d\n", res);

    return 0;
}

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