题意

题目链接

Sol

比较套路吧，设\(f[i][j]\)表示以\(i\)为根的子树中选了\(j\)个黑点对答案的贡献

然后考虑每条边的贡献，边的两边的答案都是可以算出来的

转移的时候背包一下。

#include<bits/stdc++.h>
#define Pair pair<int, int>
#define fi first
#define se second
#define MP(x, y) make_pair(x, y)
#define LL long long 
const int MAXN = 2001, INF = 1e9 + 7;
using namespace std;
inline int read() {
    int x = 0, f = 1; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, K, siz[MAXN];
LL f[MAXN][MAXN];
vector<Pair> v[MAXN];
void dfs(int x, int fa) {
    siz[x] = 1; f[x][1] = f[x][0] = 0;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i].fi, w = v[x][i].se;
        if(to == fa) continue;
        dfs(to, x);
        siz[x] += siz[to];
    }
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i].fi, w = v[x][i].se;
        if(to == fa) continue;
        for(int j = min(siz[x], K); j >= 0; j--) 
            for(int k = 0; k <= min(siz[to], j); k++)
                if(f[x][j - k] >= 0)
                f[x][j] = max(f[x][j], f[x][j - k] + f[to][k] + 1ll * k * (K - k) * w + 1ll * (siz[to] - k) *  (N - (K - k) - siz[to]) * w);
    }
}
 main() {
    N = read(); K = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read(), w = read();
        v[x].push_back(MP(y, w));
        v[y].push_back(MP(x, w));
    }
    memset(f, -0x7f, sizeof(f));
    dfs(1, 0);
    cout << f[1][K];
    return 0;
}