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 + ;

 struct edge {
     int next, to, v;
     edge(int _n = , int _t = , int _v = ) : next(_n), to(_t), v(_v) {}
 } e[N << ];

 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] = ;
     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 = ; i < n; ++i) {
         scanf("%d%d%d", &x, &y, &z);
         Add_Edges(x, y, z);
     }
     dfs(, , );
     printf("%lld\n", f[][k]);
     return ;
 }