Pre

想不出来，不得不说这套路还不错。

调试代码到了最后发现有一个地方 \(u\) 写成了 \(to[i]\)

Solution

考虑每一条边的贡献。

\(w=w_i*cntblack_{left}*cntblack_{right}+w_i*cntwhite_{left}*cntwhite_{right}\)

以这个等式进行 \(DP\)

最终计算答案。

Code

#include <cstdio>
#include <cstring>
#include <limits.h>
#define ll long long
using namespace std;
inline ll min (ll u, ll v) { return u > v ? v : u;}
inline ll max (ll u, ll v) { return u > v ? u : v;}
const int N = 2000 + 5;
int n, k;
int fr[N << 1], to[N << 1], h[N], tot, sz[N];
ll val[N << 1], dp[N][N];
inline void dfs (int u, int f) {
    memset (dp[u], -1, sizeof dp[u]);
    dp[u][0] = dp[u][1] = 0;
    sz[u] = 1;
    for (int i = h[u]; i; i = fr[i]) {
        if (to[i] == f) continue;
        dfs (to[i], u);
        sz[u] += sz[to[i]];
    }
    for (int i = h[u]; i; i = fr[i]) {
        if (to[i] == f) continue;
        for (int c = min (k, sz[u]); c >= 0; --c) {
            for (int j = 0; j <= min (c, sz[to[i]]); ++j) {
                if (~dp[u][c - j])
                    dp[u][c] = max (dp[u][c], dp[u][c - j] + dp[to[i]][j] + val[i] * j * (k - j) + val[i] * (sz[to[i]] - j) * (n - k + j - sz[to[i]]));
            }
        }
    }
}
inline void add (int u, int v, ll w) {
    tot++;
    fr[tot] = h[u];
    to[tot] = v;
    h[u] = tot;
    val[tot] = w;
}
int main () {
    #ifdef chitongz
    freopen ("x.in", "r", stdin);
    #endif
    scanf ("%d%d", &n, &k);
    for (int i = 1; i < n; ++i) {
        int x, y; ll z;
        scanf ("%d%d%lld", &x, &y, &z);
        add (x, y, z); add (y, x, z);
    }
    dfs (1, 0);
    printf ("%lld\n", dp[1][k]);
    return 0;
}

Conclusion

尝试换一个方向思考。