제출 #1247459

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1247459		MateiKing80	Chase (CEOI17_chase)	C++20	0 / 100	376 ms	181004 KiB

chase

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
#define int ll

const int N = 1e5 + 5, V = 105, inf = 1e9;

int dp1[V][N], dp2[V][N], p[N], v, ans; //<- sper ca am destula memorie
vector<int> vec[N];

void dfs(int nod, int papa) {
	for (auto i : vec[nod])
		if (i != papa)
			dfs(i, nod);
	int sumvec = p[papa], mx = -inf;
	vector<vector<int>> dpp(2, vector<int>(v + 1, -inf));
	for (auto i : vec[nod])
		if (i != papa) {
			mx = max(mx, dp2[v - 1][i] - p[i]);
			sumvec += p[i];
			for (int j = 0; j <= v; j ++) {
				if (dpp[0][j] < dp1[j][i] - p[i]) {
					dpp[1][j] = dpp[0][j];
					dpp[0][j] = dp1[j][i] - p[i];
				} else if (dpp[1][j] < dp1[j][i] - p[i])
					dpp[1][j] = dp1[j][i] - p[i];
			}
		}
	/*
	avem 3 variante:
	incepe din nod, se termina in nod sau contine nod undeva random
	*/
	ans = max(ans, dpp[0][v - 1] + sumvec);
	ans = max(ans, mx + sumvec + p[nod]);
	for (auto i : vec[nod]) {
		if (i == papa)
			continue;
		for (int j = 1; j < v - 1; j ++) {
			int x = v - 1 - j;
			ans = max(ans, dp2[j][i] - p[i] + p[nod] + sumvec + ((dp1[x][i] - p[i] == dpp[0][x]) ? dpp[1][x] : dpp[0][x]));
		}
	}
	
	for (auto i : vec[nod]) {
		if (i != papa) {
			for (int j = 1; j < v; j ++)
				dp1[j + 1][nod] = max(dp1[j + 1][nod], dp1[j][i] + sumvec - p[papa] + p[nod]);
			for (int j = 1; j < v; j ++)
				dp2[j + 1][nod] = max(dp2[j + 1][nod], dp2[j][i] + sumvec - p[papa] + p[nod]);
		}
	}
	dp1[1][nod] = sumvec - p[papa] + p[nod];
	dp2[1][nod] = sumvec - p[papa];
	
	for (int i = 1; i <= v; i ++) {
		dp1[i][nod] = max(dp1[i][nod], dp1[i - 1][nod]);
		dp2[i][nod] = max(dp2[i][nod], dp2[i - 1][nod]);
	}
}

signed main() {
	int n;
	cin >> n >> v;
	for (int i = 1; i <= n; i ++)
		cin >> p[i];
	memset(dp1, -inf, sizeof(dp1));
	memset(dp2, -inf, sizeof(dp2));
	for (int i = 1; i < n; i ++) {
		int x, y;
		cin >> x >> y;
		vec[x].push_back(y);
		vec[y].push_back(x);
	}
	dfs(1, 0);
	cout << ans << '\n';
}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 20
• Subtask 3: 0 / 30
• Subtask 4: 0 / 30