BZOJ4033 [HAOI2015]T1

令$f[p][i]$表示以$p$为根的子树内，选了$i$个黑点，剩下的都是白点的这个子树内贡献的答案

如果$p$的子树都算出来了，只要计算$p$与$fa[p]$之间的边对答案的贡献就好了，贡献是$dis * (i * (sz - i) + (k - i) * (n - k - (sz - i)))$

于是树形动规一下就好了

/**************************************************************
    Problem: 4033
    User: rausen
    Language: C++
    Result: Accepted
    Time:308 ms
    Memory:32300 kb
****************************************************************/
 
#include <cstdio>
#include <cstring>
#include <algorithm>
 
using namespace std;
typedef long long ll;
 
const int N = 2e3 + 5;
 
struct edge {
    int next, to, v;
    edge(int _n = 0, int _t = 0, int _v = 0) : next(_n), to(_t), v(_v) {}
} e[N << 1];
 
int n, k;
int first[N], tot;
int sz[N];
ll f[N][N], tmp[N];
 
inline void Add_Edges(int x, int y, int z) {
    e[++tot] = edge(first[x], y, z), first[x] = tot;
    e[++tot] = edge(first[y], x, z), first[y] = tot;
}
 
inline ll calc(int x, int y) {
    return 1ll * y * (x - y);
}
 
#define y e[x].to
inline void dfs(int p, int fa, int w) {
    int x, i, j;
    sz[p] = 1;
    for (x = first[p]; x; x = e[x].next)
        if (y != fa) {
            dfs(y, p, e[x].v);
            memcpy(tmp, f[p], sizeof(tmp));
            for (i = min(sz[p], k); ~i; --i)
                for (j = min(sz[y], k - i); ~j; --j)
                    tmp[i + j] = max(tmp[i + j], f[p][i] + f[y][j]);
            memcpy(f[p], tmp, sizeof(tmp));
            sz[p] += sz[y];
        }
    for (i = min(sz[p], k); ~i; --i)
        f[p][i] += (calc(k, i) + calc(n - k, sz[p] - i)) * w;
}
#undef y
 
int main() {
    int i, x, y, z;
    scanf("%d%d", &n, &k);
    for (i = 1; i < n; ++i) {
        scanf("%d%d%d", &x, &y, &z);
        Add_Edges(x, y, z);
    }
    dfs(1, 0, 0);
    printf("%lld
", f[1][k]);
    return 0;
}