Submission #77477

#	Time	Username	Problem	Language	Result	Execution time	Memory
77477		qkxwsm	Beads and wires (APIO14_beads)	C++17	100 / 100	257 ms	19984 KiB

beads

/*
    _____
  .'     '.
 /  0   0  \
|     ^     |
|  \     /  |
 \  '---'  /
  '._____.'
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

struct chash
{
	int operator()(int x) const
	{
		x ^= (x >> 20) ^ (x >> 12);
		return x ^ (x >> 7) ^ (x >> 4);
	}
};

template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename U> using hashtable = gp_hash_table<T, U, chash>;
random_device(rd);
mt19937 rng(rd());

template<class T>
void readi(T &x)
{
	T input = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		input = input * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		input = -input;
	}
	x = input;
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int aout[20];
	int ilen = 0;
	while(output)
	{
		aout[ilen] = ((output % 10));
		output /= 10;
		ilen++;
	}
	for (int i = ilen - 1; i >= 0; i--)
	{
		putchar(aout[i] + '0');
	}
	return;
}
template<class T>
void ckmin(T &a, T b)
{
	a = min(a, b);
}
template<class T>
void ckmax(T &a, T b)
{
	a = max(a, b);
}
template<class T, class U>
T nmod(T &x, U mod)
{
	if (x >= mod) x -= mod;
}
template<class T>
T gcd(T a, T b)
{
	return (b ? gcd(b, a % b) : a);
}
template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define MT make_tuple
#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define debug(x) cerr << #x << " = " << x << endl;

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-10;

#define MAGIC 347
#define SINF 10007
#define CO 1000007
#define INF 1000000007
#define BIG 1000000931
#define LARGE 1696969696967ll
#define GIANT 2564008813937411ll
#define LLINF 2696969696969696969ll
#define MAXN 200013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;

int N;
vector<pii> edge[MAXN];
int parent[MAXN], dup[MAXN];
vector<int> ord;
int res[2][2];
int dp[MAXN][2][2];
int mx0 = -INF, mx1 = -INF;

void dfs(int u)
{
	for (pii p : edge[u])
	{
		int d = p.fi, v = p.se;
		if (v == parent[u])
		{
			continue;
		}
		parent[v] = u;
		dup[v] = d;
		dfs(v);
	}
	ord.PB(u);
}
void init()
{
	res[0][0] = 0; res[0][1] = 0; res[1][0] = 0; res[1][1] = 0;
	mx0 = -INF; mx1 = -INF;
}
void assign(int u)
{
	ckmax(dp[u][0][0], res[0][0]);
	ckmax(dp[u][0][1], res[0][1]);
	ckmax(dp[u][1][0], res[1][0]);
	ckmax(dp[u][1][1], res[1][1]);
}

int32_t main()
{
	ios_base::sync_with_stdio(0);
	// cout << fixed << setprecision(10);
	// cerr << fixed << setprecision(10);
	// freopen ("file.in", "r", stdin);
	// freopen ("file.out", "w", stdout);
	cin >> N;
	for (int i = 1; i < N; i++)
	{
		int u, v, d;
		cin >> u >> v >> d;
		u--; v--;
		edge[u].PB({d, v});
		edge[v].PB({d, u});
	}
	parent[0] = N; dup[0] = -INF;
	dfs(0);
	for (int i = 0; i < N; i++)
	{
		int u = ord[i];
		// cerr << u << endl;
		//dp[vertex][taking only guy to parent][have u gotten a cut]
		// res[1][0] = dup[u] + sum over all v of dp[v][0][0]
		// res[1][1] = dup[u] + sum over all v of dp[v][0][1]
		for (pii p : edge[u])
		{
			int v = p.se;
			if (v == parent[u]) continue;
			res[0][0] += dp[v][0][0];
			res[0][1] += dp[v][0][0];
			ckmax(mx1, dp[v][0][1] - dp[v][0][0]);
		}
		res[0][1] += mx1;
		res[1][0] = res[0][0] + dup[u];
		res[1][1] = res[0][1] + dup[u];
		assign(u); init();
		//if u dont take u, then:
		//res[0][0] = sum over all v of dp[v][0][0]
		//res[0][1] = sum over all v of dp[v][0][1]
		//if u take two chidlren of u: it's always ok
		//res[0][1] = sum over all v except 2 of dp[v][0][1] plus the two dp[v][1]][1]
		int sum = 0;
		//u choose vertices v1, v2 and v3 and take dp[v1][1][0] + dp[v2][1][1] + dp[rest][0][0]
		for (pii p : edge[u])
		{
			int v = p.se;
			if (v == parent[u]) continue;
			int dif = dp[v][1][0] - dp[v][0][0];
			if (dif > mx0)
			{
				mx1 = mx0; mx0 = dif;
			}
			else if (dif > mx1)
			{
				mx1 = dif;
			}
			sum += dp[v][0][0];
		}
		for (pii p : edge[u])
		{
			int v = p.se;
			if (v == parent[u]) continue;
			ckmax(res[0][1], (dp[v][1][1] - dp[v][0][0]) + ((dp[v][1][0] - dp[v][0][0] == mx0) ? mx1 : mx0) + sum);
		}
		res[1][1] = res[0][1] + dup[u];
		assign(u); init();
		//if u take one child of u and v
		//res[0][0] = sum over all v except 1 of dp[v][0][0] plus dp[v][1][0] + dup[u]
		//res[0][1] = sum over all v except 1 of dp[v][0][0] plus dp[v][1][1] + dup[u]
		res[0][0] = dup[u];
		res[0][1] = dup[u];
		for (pii p : edge[u])
		{
			int v = p.se;
			if (v == parent[u]) continue;
			res[0][0] += dp[v][0][0];
			res[0][1] += dp[v][0][0];
			ckmax(mx0, dp[v][1][0] - dp[v][0][0]);
			ckmax(mx1, dp[v][1][1] - dp[v][0][0]);
		}
		res[0][0] += mx0; res[0][1] += mx1;
		assign(u); init();
		//u just take nothing
		ckmax(dp[u][1][1], dp[u][1][0]);
		ckmax(dp[u][0][1], dp[u][0][0]);
		// cerr << "vertex " << u << ' ' << dp[u][0][0] << ' ' << dp[u][0][1] << ' ' << dp[u][1][0] << ' ' << dp[u][1][1] << endl;
	}
	cout << dp[0][0][1] << '\n';
	//possibilities: u took 1 or u havent
	//ur subtree is stupid or its not
	//the path between anytwo endpoints must go through a third point?
	// cerr << "time elapsed = " << (clock() / (CLOCKS_PER_SEC / 1000)) << " ms" << endl;
	return 0;
}

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