由於是中文題目,就不說題意了。直接思路:樹形DP, 定義狀態dp[i][j]表示以i爲根的子樹中選j個點(i點必選)所能獲得最高效益,我新建一個0號節點連接所有直接能獲取寶貝的點,這樣最後答案就是dp[0][m+1]。轉移方程就是類似01揹包的轉移,dp[u][k] = max(dp[u][k], dp[u][k-j]+dp[v][j]).
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define N 202
int val[N], m, n, dp[N][N], tmp[N];
vector<int>V[N];
void dfs(int u) {
int i, v, j, k;
// printf("%d\n", u);
dp[u][1] = val[u], dp[u][0] = 0;
for (i = 0;i < V[u].size();i++) {
v = V[u][i];
dfs(v);
memcpy(tmp, dp[u], sizeof(tmp));
for (k = m+1;k >= 1;k--) {
for (j = 0;j < k;j++) {
if (dp[v][j] == -1) continue;
if (dp[u][k-j] == -1) continue;
if (dp[u][k] == -1) dp[u][k] = dp[u][k-j]+dp[v][j];
dp[u][k] = max(dp[u][k], dp[u][k-j]+dp[v][j]);
}
}
}
}
int main() {
int a, b, i, j;
while (scanf("%d%d", &n, &m), n) {
for (i = 0;i <= n;i++) V[i].clear();
val[0] = 0;
memset(dp, -1, sizeof(dp));
for (i = 1;i <= n;i++) {
scanf("%d%d", &a, &val[i]);
V[a].push_back(i);
}
dfs(0);
printf("%d\n", dp[0][m+1]);
}
}