제출 #218902

#제출 시각아이디문제언어결과실행 시간메모리
218902eggag32Putovanje (COCI20_putovanje)C++17
0 / 110
211 ms24820 KiB
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) debug(x) debug(y);
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end() 
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))

template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }

int vis1[200005]; //don't forget to initialize to zeroes
vi ord; //don't forget to reverse in the end
vi g[200005];
 
void dfs0(int cur){
	vis1[cur] = 1;
	for(int x : g[cur]) if(!vis1[x]) dfs0(x);
	ord.pb(cur);
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	//freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
	int n;
	cin >> n;
	map<pi, pi> m;
	rep(i, n - 1){
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		a--, b--;
		g[a].pb(b);
		g[b].pb(a);
		m[mp(a, b)] = mp(c, d);
		m[mp(b, a)] = mp(c, d);
	}
	memset(vis1, 0, sizeof(vis1));
	rep(i, n) if(!vis1[i]) dfs0(i);
	reverse(all(ord));
	vi pos(n, 0);
	rep(i, n) pos[ord[i]] = i;
	vi cnt(n + 5, 0);
	repn(i, 1, n){
		//we go from 0 -> 1, 1 -> 2, 2 -> 3
		if(pos[i - 1] < pos[i]) cnt[pos[i - 1] + 1]++, cnt[pos[i] + 1]--;
		else cnt[pos[i] + 1]++, cnt[pos[i - 1] + 1]--;
	}
	partial_sum(all(cnt), cnt.begin());
	ll ans = 0;
	repn(i, 1, n){
		pi x = m[mp(ord[i - 1], ord[i])];
		ans += min(cnt[i] * x.fi, x.se);
	}
	cout << ans << endl;
	return 0;
}
/*
Things to look out for:
	- Integer overflows
	- Array bounds
	- Special cases
Be careful!
*/
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