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ABC-394-F-Alkane(烷烃)

思路讲解

这个下面的题目好像是问怎么把一个图转化为雪花树

ABC-385-E - Snowflake Tree

这个是要问Alkane的最大子图(边数最多)

给的无向图是一颗树。

“烷烃”的定义是这颗树上,顶点的入度为4或者为1,而且必须有一个点的入度为4

其实我感觉这种题目,这种比较简单的树上dp没什么难的,就是要分类讨论,每种情况都要想清楚。

AC代码

https://atcoder.jp/contests/abc394/submissions/64007396

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// Problem: F - Alkane
// Contest: AtCoder - KAJIMA CORPORATION CONTEST 2025 (AtCoder Beginner Contest 394)
// URL: https://atcoder.jp/contests/abc394/tasks/abc394_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(2e5)+10,INF=static_cast<ll>(5e18)+3;
typedef pair<ll,bool> plb;

ll N,M,T,A[MAXN];
vector<ll> g[MAXN];
ll ans=0;
bool vis[MAXN];
ll dp[MAXN];

pair<ll,bool> dfs(ll x){
ll lans=0;
bool flag=false;
vector<ll> rett;
FOR(i,0,SZ(g[x])-1){
ll node=g[x][i];
if(!vis[node]){
vis[node]=true;
plb ret=dfs(node);
if(ret.se) flag=true;
lans+=ret.fi;
rett.pb(ret.fi);
}
}
if(SZ(g[x])==4){
lans+=1;
ans=max(lans,ans);
return (plb){lans,true};
}else if(SZ(g[x])==1){
lans+=1;
if(flag) ans=max(lans,ans);
return (plb){lans,flag};
}else if(SZ(g[x])<4){
if(flag){
sort(all(rett));
ans=max(ans,1+rett[SZ(rett)-1]);
}
return (plb){1,false};
}else{
sort(all(rett));
lans=0;
ROF(i,SZ(rett)-1,SZ(rett)-3){
lans+=rett[i];
}
lans+=1;
ans=max(ans,lans+rett[SZ(rett)-4]);
return (plb){lans,true};
}
}

inline void solve(){
cin>>N;
ans=0;
FOR(i,1,N-1){
ll a,b;
cin>>a>>b;
g[a].pb(b);
g[b].pb(a);
}
vis[1]=true;
dfs(1);
cout<<(ans==0?-1:ans)<<"\n";
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
solve();
return 0;
}
/*
WA
http://atcoder.jp/contests/abc394/submissions/64001178
AC
https://atcoder.jp/contests/abc394/submissions/64007396
*/

心路历程(WA,TLE,MLE……)

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7
1 2
2 3
3 4
3 5
3 6
3 7

ans: 5

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2 3
3 4
3 5
5 6
5 7
5 8

ans:5
hack: https://atcoder.jp/contests/abc394/submissions/64007220
该提交在入度为2or3的处理上有一点问题

hack数据以上

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// Problem: F - Alkane
// Contest: AtCoder - KAJIMA CORPORATION CONTEST 2025 (AtCoder Beginner Contest 394)
// URL: https://atcoder.jp/contests/abc394/tasks/abc394_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
constexpr ll MAXN=static_cast<ll>(2e5)+10,INF=static_cast<ll>(5e18)+3;
typedef pair<ll,bool> plb;

ll N,M,T,A[MAXN];
vector<ll> g[MAXN];
ll ans=0;
bool vis[MAXN];

pair<ll,bool> dfs(ll x){
ll lans=0;
bool flag=false;
FOR(i,0,SZ(g[x])-1){
ll node=g[x][i];
if(!vis[node]){
vis[node]=true;
plb ret=dfs(node);
if(ret.se) flag=true;
lans+=ret.fi;
}
}
if(SZ(g[x])==4){
lans+=1;
#ifdef LOCAL
cerr << x<<" "<<lans << '\n';
#endif
ans=max(lans,ans);
return (plb){lans,true};
}else if(SZ(g[x])==1){
lans+=1;
#ifdef LOCAL
cerr << x<<" "<<lans << '\n';
#endif
if(flag) ans=max(lans,ans);
return (plb){lans,flag};
}else{
return (plb){1,false};
}
}

inline void solve(){
cin>>N;
ans=0;
FOR(i,1,N-1){
ll a,b;
cin>>a>>b;
g[a].pb(b);
g[b].pb(a);
}
vis[1]=true;
dfs(1);
cout<<(ans==0?-1:ans)<<"\n";
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
solve();
return 0;
}
/*
WA
http://atcoder.jp/contests/abc394/submissions/64001178
*/