Submission #670215

#	Time	Username	Problem	Language	Result	Execution time	Memory
670215		MohammadAghil	Beads and wires (APIO14_beads)	C++17	0 / 100	14 ms	23792 KiB

beads

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
 using namespace std;
  typedef long long ll;
   typedef pair<ll, ll> pp;
     #define per(i,r,l) for(int i = (r); i >= (l); i--)
       #define rep(i,l,r) for(int i = (l); i < (r); i++)
          #define all(x) begin(x), end(x)
             #define sz(x) (int)(x).size()
                 #define pb push_back
                     #define ss second
                          #define ff first
                                  void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
void IOS(){
     cin.tie(0) -> sync_with_stdio(0);
     #ifndef ONLINE_JUDGE
          #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
     #else
          #define er(args ...) 0
     #endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e6 + 5, maxk = 2e2 + 5, p = 9973, lg = 22, inf = ll(1e18) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll,md);return k*k%md*(b&1ll?a:1)%md;}

ll dp[maxn][2];
vector<pp> adj[maxn];

void dfs(int r, int p, int vl){
     ll sum = 0;
     pp mx = {-inf, -inf};
     for(auto[c, w]: adj[r]) if(c - p){
          dfs(c, r, w);
          sum += dp[c][1];
          ll v = w - dp[c][1] + dp[c][0];
          if(mx.ff < v) mx = {v, mx.ff};
          else if(mx.ss < v) mx.ss = v;
     }     
     dp[r][0] = max(sum, sum + mx.ff + mx.ss);
     dp[r][1] = max(dp[r][0], sum + mx.ff + vl);
     // er(r+1, dp[r][0], dp[r][1]);
}

int main(){ IOS();     
     int n; cin >> n;
     rep(i,1,n){
          int u, v, w; cin >> u >> v >> w; u--, v--;
          adj[u].pb({v, w});
          adj[v].pb({u, w});  
     }
     dfs(0, -1, 0);
     cout << dp[0][0] << '\n';
     return 0-0;
}

• Subtask 1: 0 / 13
• Subtask 2: 0 / 15
• Subtask 3: 0 / 29
• Subtask 4: 0 / 43