Submission #110044

#	Time	Username	Problem	Language	Result	Execution time	Memory
110044		qkxwsm	Election Campaign (JOI15_election_campaign)	C++14	100 / 100	525 ms	28784 KiB

election_campaign

//clever
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

random_device(rd);
mt19937 rng(rd());
const long long FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();

struct custom_hash
{
	template<class T>
	unsigned long long operator()(T v) const
	{
		unsigned long long x = v;
		x += FIXED_RANDOM;
		// x += 11400714819323198485ull;
		// x = (x ^ (x >> 30)) * 13787848793156543929ull;
		x = (x ^ (x >> 27)) * 10723151780598845931ull;
		return x ^ (x >> 31);
	}
};

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U> using hash_table = gp_hash_table<T, U, custom_hash>;

template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}
template<class T>
void readi(T &x)
{
	x = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		x = x * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		x = -x;
	}
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int buf[20], n = 0;
	while(output)
	{
		buf[n] = ((output % 10));
		output /= 10;
		n++;
	}
	for (n--; n >= 0; n--)
	{
		putchar(buf[n] + '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);
}
long long expo(long long a, long long e, long long mod)
{
	return ((e == 0) ? 1 : ((expo(a * a % mod, e >> 1, mod)) * ((e & 1) ? a : 1) % mod));
}
template<class T, class U>
void 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);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define DBG(x) cerr << #x << " = " << (x) << endl
#define SZ(x) ((int) ((x).size()))
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) (x).begin(), (x).end()

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-9;

#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 100013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<pdd> vpd;

int N, M, T;
vi edge[MAXN];
vector<pair<int, pii> > choice[MAXN];
int parent[MAXN], st[MAXN], ft[MAXN], depth[MAXN];
int ancestor[25][MAXN];
int dp[MAXN], sdp[MAXN];
vi ord;

int fen[MAXN];
void upd(int idx, int v)
{
	for (int e = idx + 1; e <= N + 1; e += e & (-e))
	{
		fen[e] += v;
	}
	return;
}
void rangeupdate(int l, int r, int v)
{
	upd(l, v);
	upd(r + 1, -v);
}
int query(int idx)
{
	int res = 0;
	for (int e = idx + 1; e; e -= e & (-e))
	{
		res += fen[e];
	}
	return res;
}
void dfs(int u)
{
	st[u] = T; ft[u] = T;
	T++;
	for (int v : edge[u])
	{
		if (v == parent[u]) continue;
		parent[v] = u;
		// cerr << "parent " << v << " = " << u << endl;
		depth[v] = depth[u] + 1;
		dfs(v);
		ft[u] = ft[v];
	}
	ord.PB(u);
}
bool insubt(int u, int v)
{
	if (u == N) return true;
	if (st[u] <= st[v] && st[v] <= ft[u]) return true;
	return false;
}
int lca(int u, int v)
{
	//how far does v need to go up to?
	if (insubt(v, u)) return -1;
	// cerr << "solver " << u << ' ' << v << " is ";
	FORD(i, 20, 0)
	{
		if (!insubt(ancestor[i][v], u)) v = ancestor[i][v];
	}
	// cerr << v << endl;
	return v;
}
void solve(int u)
{
	for (int v : edge[u])
	{
		if (v == parent[u]) continue;
		sdp[u] += dp[v];
	}
	int res = 0;
	for (auto p : choice[u])
	{
		int cur = p.fi;
		cur += query(st[p.se.fi]);
		cur += query(st[p.se.se]);
		// int v = p.se.fi;
		// while(v != u)
		// {
		// 	cur += (sdp[v] - dp[v]);
		// 	v = parent[v];
		// }
		// v = p.se.se;
		// while(v != u)
		// {
		// 	cur += (sdp[v] - dp[v]);
		// 	v = parent[v];
		// }
		ckmax(res, cur);
	}
	res += sdp[u];
	dp[u] = res;
	rangeupdate(st[u], ft[u], sdp[u] - dp[u]);
}

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	// cout << fixed << setprecision(10);
	// cerr << fixed << setprecision(10);
	// if (fopen("file.in", "r"))
	// {
	// 	freopen ("file.in", "r", stdin);
	// 	freopen ("file.out", "w", stdout);
	// }
	cin >> N;
	FOR(i, 0, N - 1)
	{
		int u, v;
		cin >> u >> v; u--; v--;
		edge[u].PB(v);
		edge[v].PB(u);
	}
	parent[0] = N;
	dfs(0);
	FOR(i, 0, 25) ancestor[i][N] = N;
	FOR(i, 0, N) ancestor[0][i] = parent[i];
	// FOR(i, 0, N)
	// {
	// 	cerr << "ancestor " << 0 << ' ' << i << " = " << parent[i] << endl;
	// }
	FOR(i, 1, 24)
	{
		FOR(j, 0, N)
		{
			ancestor[i][j] = ancestor[i - 1][ancestor[i - 1][j]];
		}
	}
	// DBG(ancestor[1][1]);
	cin >> M;
	FOR(i, 0, M)
	{
		int u, v, c;
		cin >> u >> v >> c; u--; v--;
		if (depth[u] > depth[v]) swap(u, v);
		int w = lca(u, v); w = parent[w];
		choice[w].PB({c, {u, v}});
		// cerr << w << " gets " << c << " with " << u << " -> " << v << endl;
	}
	for (int u : ord)
	{
		solve(u);
	}
	cout << dp[0] << '\n';
	// cerr << "time elapsed = " << (clock() / (CLOCKS_PER_SEC / 1000)) << " ms" << endl;
	return 0;
}
/* READ READ READ
* int overflow, maxn too small, special cases (n=1?, two distinct?), cin.tie() interactive
* reread the problem, try small cases
* note down possible sources of error as you go
* do smth instead of nothing
*/

• Subtask 1: 10 / 10
• Subtask 2: 5 / 5
• Subtask 3: 5 / 5
• Subtask 4: 30 / 30
• Subtask 5: 10 / 10
• Subtask 6: 40 / 40