제출 #305483

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
305483		MarcoMeijer	구슬과 끈 (APIO14_beads)	C++14	28 / 100	1083 ms	3072 KiB

beads

#include <bits/stdc++.h>
using namespace std;
 
// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e18
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }
 
// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const vector<T>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi,' ',x.se);}
template<class T> void OUT(const vector<T>& x) {RE(i,x.size()) OUT(i==0?"":" ",x[i]);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
template<class H> void OUTLS(const H& h) {OUTL(h); }
template<class H, class... T> void OUTLS(const H& h, const T&... t) {OUT(h,' '); OUTLS(t...); }
 
//===================//
//  Added libraries  //
//===================//
 
//===================//
//end added libraries//
//===================//
 
void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 
 
//===================//
//   begin program   //
//===================//
 
const int MX = 1e5+2;

int n;
vii adj[MX];
int dp[MX][2];
int root;

void dfsDP(int u, int p=-1) {
    FOR(v,adj[u]) if(v.fi!=p) dfsDP(v.fi, u);

    dp[u][0] = 0;
    FOR(v,adj[u]) if(v.fi!=p) dp[u][0] += max(dp[v.fi][1]+v.se, dp[v.fi][0]);

    dp[u][1] = -1e9;
    int sm = 0;
    FOR(v,adj[u]) if(v.fi!=p) sm += max(dp[v.fi][1]+v.se, dp[v.fi][0]);
    FOR(v,adj[u]) if(v.fi!=p) {
        dp[u][1] = max(dp[u][1], sm - max(dp[v.fi][1]+v.se, dp[v.fi][0]) + dp[v.fi][0] + v.se);
    }
}

void program() {
    IN(n);
    RE(i,n-1) {
        int u, v, w;
        IN(u,v,w);
        adj[u].pb({v,w});
        adj[v].pb({u,w});
    }

    root = 1;
    RE1(i,n) if(adj[i].size() == 1) root = i;

    int bst = 0;
    RE1(i,n) {
        root = i;
        if(adj[i].size() != 1) continue;
        dfsDP(root);
        bst = max(bst, dp[root][0]);
    }
    OUTL(bst);
}

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