제출 #151821

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
151821		qkxwsm	친구 (IOI14_friend)	C++14	35 / 100	6 ms	2808 KiB

friend

#include "friend.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
	if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
	if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#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 SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define MAXN 100013

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

int N;
int arr[MAXN];
int dsu[MAXN];
int val[MAXN];
vi edge[MAXN];
int dp[MAXN][2];
int ans;

int get(int u)
{
	return (u == dsu[u] ? u : dsu[u] = get(dsu[u]));
}
void merge(int u, int v)
{
	u = get(u); v = get(v);
	dsu[u] = v;
	return;
}
void solve(int u, int p)
{
	dp[u][0] = 0;
	dp[u][1] = val[u];
	for (int v : edge[u])
	{
		if (v == p) continue;
		solve(v, u);
		dp[u][0] += dp[v][1];
		dp[u][1] += dp[v][0];
	}
	ckmax(dp[u][1], dp[u][0]);
}

int findSample(int n, int confidence[], int host[], int protocol[])
{
	N = n;
	FOR(i, 0, N)
	{
		arr[i] = confidence[i];
		dsu[i] = i;
	}
	//0 im: just a normal edge. 1 myf: they literally merge. 2 wereyourfriends:
	//so you somehow want to store like components and info about them.
	FORD(i, N, 1)
	{
		int p = host[i], qid = protocol[i];
		if (qid == 1)
		{
			merge(i, p);
		}
	}
	FOR(i, 0, N)
	{
		val[get(i)] += arr[i];
	}
	FOR(i, 0, N)
	{
		arr[i] = val[i];
	}
	//the structure is just like, you have a bunch of cliques inside a tree
	FORD(i, N, 1)
	{
		int p = host[i], qid = protocol[i];
		if (qid == 2)
		{
			//we are your friends!
			merge(i, p);
		}
	}
	FOR(i, 0, N)
	{
		ckmax(val[get(i)], arr[i]);
	}
	FORD(i, N, 1)
	{
		int p = host[i], qid = protocol[i];
		if (qid == 0)
		{
			//yeah
			edge[get(p)].PB(get(i));
			edge[get(i)].PB(get(p));
		}
	}
	FOR(i, 0, N)
	{
		if (get(i) == i)
		{
			solve(i, N);
			ans = dp[i][1];
			break;
		}
	}
	return ans;
}

• Subtask 1: 0 / 11
• Subtask 2: 8 / 8
• Subtask 3: 8 / 8
• Subtask 4: 19 / 19
• Subtask 5: 0 / 23
• Subtask 6: 0 / 31