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답안 #791262

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
791262 2023-07-23T20:23:16 Z dungz Dostavljač (COCI18_dostavljac) C++17
140 / 140
 201 ms 2500 KB 
// COCI 2017-2018 - Dostavljac
// Lúcio Cardoso

// Solution:

// 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.

// 2. At first, dp[0][u][k] = dp[1][u][k] = A[k], for all k > 0. Then, we can use "knapsack on trees".
// For each child v of u, we have:

// * dp[0][u][k] = max(dp[0][u][k], max(dp[0][v][j], dp[1][v][j]) + dp[1][u][i-j-1]), for 0 <= j < k (we end our path inside the subtree of v)
// * dp[0][u][k] = max(dp[0][u][k], dp[1][v][j] + dp[0][u][i-j-2]), for 0 <= j < k-1 (we end our path inside the subtree of another child)
// * dp[1][u][k] = max(dp[1][u][k], dp[1][v][j] + dp[1][u][i-j-2]), for 0 <= j < k-1 (we can go through v's subtree)

// 3. Final answer is max(dp[0][1][m], dp[1][1][m]). Overall complexity if O(n*m^2).

#include <bits/stdc++.h>

using namespace std;

const int maxn = 510;

int n, m;
int A[maxn];

// 0 - ending inside subtree
// 1 - coming back to vertex
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

dostavljac.cpp: In function 'int main()':
dostavljac.cpp:63:7: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   63 |  scanf("%d %d", &n, &m);
      |  ~~~~~^~~~~~~~~~~~~~~~~
dostavljac.cpp:66:8: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   66 |   scanf("%d", &A[i]);
      |   ~~~~~^~~~~~~~~~~~~
dostavljac.cpp:71:8: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
   71 |   scanf("%d %d", &u, &v);
      |   ~~~~~^~~~~~~~~~~~~~~~~

Subtask #1
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 396 KB Output is correct

Subtask #2
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 340 KB Output is correct

Subtask #3
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 468 KB Output is correct

Subtask #4
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 580 KB Output is correct

Subtask #5
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 676 KB Output is correct
2 Correct 2 ms 724 KB Output is correct

Subtask #6
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 836 KB Output is correct
2 Correct 30 ms 920 KB Output is correct

Subtask #7
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 54 ms 1084 KB Output is correct
2 Correct 16 ms 1132 KB Output is correct

Subtask #8
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 18 ms 1448 KB Output is correct
2 Correct 120 ms 1524 KB Output is correct

Subtask #9
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 101 ms 1912 KB Output is correct
2 Correct 41 ms 1908 KB Output is correct

Subtask #10
14.0 / 14.0

# 결과 실행 시간 메모리 Grader output
1 Correct 201 ms 2500 KB Output is correct
2 Correct 19 ms 2252 KB Output is correct