# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
1161906 | mnbvcxz123 | Dostavljač (COCI18_dostavljac) | C++20 | 269 ms | 2420 KiB |
// 1. Let dp[0][u][k] be the greatest amount we can achieve by starting our path at vertex u and
// finishing it inside the subtree of u, using a cost k. Similarly, let dp[1][u][k] be the same as
// the greatest amount in a path starting at u and ending at u with a cost of k.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 510;
int n, m;
int A[maxn];
int dp[2][maxn][maxn];
vector<int> grafo[maxn];
void dfs(int u, int p)
{
for (int i = 1; i <= m; i++)
dp[0][u][i] = dp[1][u][i] = A[u];
for (auto v: grafo[u])
{
if (v == p) continue;
dfs(v, u);
for (int i = m; i >= 1; i--)
{
for (int j = 0; j <= i; j++)
{
if (i >= j+1)
dp[0][u][i] = max(dp[0][u][i], max(dp[0][v][j], dp[1][v][j]) + dp[1][u][i-j-1]);
if (i >= j+2)
{
dp[0][u][i] = max(dp[0][u][i], dp[1][v][j] + dp[0][u][i-j-2]);
dp[1][u][i] = max(dp[1][u][i], dp[1][v][j] + dp[1][u][i-j-2]);
}
}
}
}
}
int main(void)
{
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &A[i]);
for (int i = 1; i < n; i++)
{
int u, v;
scanf("%d %d", &u, &v);
grafo[u].push_back(v); grafo[v].push_back(u);
}
dfs(1, 0);
printf("%d\n", max(dp[0][1][m], dp[1][1][m]));
}
Compilation message (stderr)
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