【刷题】【记忆化搜索】【dp】Longtail Hedgehog

太巧的记忆化了，考虑到尾巴要求单调递增，且要求尾巴最长，则后者状态可由前者推导

https://codeforces.com/group/oqKDf9Vyex/contest/349763/problem/D

#include<cstdio>
#include<cstdlib>
#include<vector>
#include<iostream>
using namespace std;
#define ll long long

int n,m;
const int N=1e5+10;
vector <int > g[N];
ll f[N],ans;//ans=len(tail)*(g[end].size()-1)

int main()
{
    cin>>n>>m;
    while(m--)
    {
        int u,v;
        cin>>u>>v;
        g[u].push_back(v); 
        g[v].push_back(u); 
    }
    
    f[1]=1,ans=g[1].size() ;
    for(int i=2;i<=n;i++)
    {
        f[i]=1;
        
        int sz=g[i].size() ;
        for(int j=0;j<sz;j++)
            if(g[i][j] < i ) f[i]=max(f[i],f[g[i][j] ]+1);
        ans=max(ans,f[i]*sz );
    }
    
    cout<<ans;
    return 0;
}