Submission #312633

#	Time	Username	Problem	Language	Result	Execution time	Memory
312633		aZvezda	Beads and wires (APIO14_beads)	C++14	0 / 100	2 ms	2688 KiB

beads

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define endl "\n"
typedef long long ll;
template<class T, class T2> inline ostream &operator <<(ostream &out, const pair<T, T2> &x) { out << x.first << " " << x.second; return out;}
template<class T, class T2> inline istream &operator >>(istream &in, pair<T, T2> &x) { in >> x.first >> x.second; return in;}
template<class T, class T2> inline bool chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline bool chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const ll mod = 1e9 + 7;
#define out(x) "{" << (#x) << ": " << x << "} "

const int MAX_N = 1e5 + 10;
int dp[MAX_N][2];
int n;
vector<pair<int, int> > g[MAX_N];

void dfs(int x, int p, int w) {
    vector<int> pos;
    int sum = 0;
    for(auto it : g[x]) {
        if(it.first == p) {continue;}
        dfs(it.first, x, it.second);
        int curr = max(dp[it.first][0], dp[it.first][1]);
        sum += curr;
        pos.push_back(dp[it.first][0] + it.second - curr);
    }

    dp[x][0] = sum;
    sort(pos.rbegin(), pos.rend());
    for(int i = 0; i < (int)pos.size() - 1; i += 2) {
        int curr = pos[i] + pos[i + 1];
        if(curr > 0) {
            dp[x][0] += curr;
        }
        break;
    }

    dp[x][1] = sum;

    pos.push_back(w);
    sort(pos.rbegin(), pos.rend());

    for(int i = 0; i < (int)pos.size() - 1; i += 2) {
        int curr = pos[i] + pos[i + 1];
        if(curr > 0) {
            dp[x][1] += curr;
        }
        break;
    }
}

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    cin >> n;
    for(int i = 0; i < n - 1; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        g[a].push_back({b, c});
        g[b].push_back({a, c});
    }
    dfs(1, 0, mod);
    cout << dp[1][0] << endl;
    return 0;
}

/*
5
1 2 10
1 3 40
1 4 15
1 5 20

10
4 10 2
1 2 21
1 3 13
6 7 1
7 9 5
2 4 3
2 5 8
1 6 55
6 8 34
*/

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