要點

• 是樹形的考慮dfs
• 分為取一枝，取兩枝兩種情況，將它們的合法情況進行暴舉取最好答案即可，貌似我亂搞得相當冗……
• 順手記憶化
• 正解應該是樹上背包

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <vector>
#define pb push_back
using namespace std;

int N, Q, cost[101][101], size[101], dp[101][101];
vector<int> adj[101];

int dfs(int cur, int fa, int rest) {
    if (rest <= 0)  return 0;
    if (dp[cur][rest] >= 0) return dp[cur][rest];

    int res = 0;
    vector<int> v;
    for (int son : adj[cur]) {
        if (son == fa)  continue;
        if (size[son] + 1 >= rest)  res = max(res, dfs(son, cur, rest - 1) + cost[cur][son]);
        v.pb(son);
    }
    if (v.size() == 2 && rest >= 2) {
        int r = rest - 2;
        for (int i = 0; i <= min(r, size[v[0]]); i++)
            if (r - i <= size[v[1]]) {
                res = max(res, dfs(v[0], cur, i) + dfs(v[1], cur, r - i) + cost[cur][v[0]] + cost[cur][v[1]]);
            }
    }
    return dp[cur][rest] = res;
}

int main() {
    scanf("%d %d", &N, &Q);
    if (Q == N) Q--;
    for (int i = 1; i < N; i++) {
        int u, v, c;
        scanf("%d %d %d", &u, &v, &c);
        cost[u][v] = cost[v][u] = c;
        adj[u].pb(v), adj[v].pb(u);
    }

    function<void(int, int)> S = [&](int cur, int fa) {
        size[cur] = 0;
        for (int son : adj[cur])
            if (son != fa) {
                S(son, cur);
                size[cur] += size[son] + 1;
            }
    };
    S(1, 0);
    memset(dp, -1, sizeof dp);

    printf("%d\n", dfs(1, 0, Q));
    return 0;
}
 

樹上背包版，j-k要使用上個兒子的所以j倒序：

#include <cstdio>
#include <algorithm>
#include <vector>
#define pb push_back
using namespace std;

int N, Q, cost[101][101], dp[101][101];
vector<int> adj[101];

int dfs(int cur, int fa) {
    int ret = 0;
    for (int son : adj[cur]) {
        if (son == fa)  continue;
        int size = dfs(son, cur);
        ret += size + 1;
        for (int j = min(Q, ret); j; j--) {
            for (int k = 1; k <= min(j, size + 1); k++) {
                dp[cur][j] = max(dp[cur][j], dp[cur][j - k] + dp[son][k - 1] + cost[cur][son]);
            }
        }
    }
    return ret;
}

int main() {
    scanf("%d %d", &N, &Q);
    if (Q == N) Q--;
    for (int i = 1; i < N; i++) {
        int u, v, c;
        scanf("%d %d %d", &u, &v, &c);
        cost[u][v] = cost[v][u] = c;
        adj[u].pb(v), adj[v].pb(u);
    }

    dfs(1, 0);
    printf("%d\n", dp[1][Q]);
    return 0;
}
 

洛谷2015（樹形dp）