bzoj 4033

树形DP，dp[i][j]表示i子树中，选了j个白点，i子树中所有边的贡献。

 /**************************************************************
     Problem: 4033
     User: idy002
     Language: C++
     Result: Accepted
     Time:6568 ms
     Memory:32492 kb
 ****************************************************************/

 #include <cstdio>
 #define min(a,b) ((a)<(b)?(a):(b))
 #define max(a,b) ((a)>(b)?(a):(b))
 #define oo 0x3f3f3f3f3f3f3f3fLL
 #define N 2010 

 typedef long long dnt; 

 int n, m;
 int head[N], next[N+N], dest[N+N], wght[N+N], etot;
 dnt siz[N];
 dnt dp[N][N], pp[N]; 

 void adde( int u, int v, int w ) {
     etot++;
     wght[etot] = w;
     dest[etot] = v;
     next[etot] = head[u];
     head[u] = etot;
 }
 void dfs( int u, int fa ) {
     siz[u] = ;
     for( int t=head[u]; t; t=next[t] ) {
         int v=dest[t];
         if( v==fa ) continue;
         dfs( v, u );
         siz[u] += siz[v];
     }
     int maxc = min( siz[u], m );
     for( int c=; c<=maxc; c++ )
         pp[c] = -oo;
     pp[] = ;
     for( int t=head[u]; t; t=next[t] ) {
         int v=dest[t], w=wght[t];
         if( v==fa ) continue;
         for( int c=maxc; c>=; c-- ) {
             int msc = min( c, siz[v] );
             for( int sc=; sc<=msc; sc++ ) {
                 dnt tv = pp[c-sc]+dp[v][sc]+((dnt)sc*(m-sc)+(siz[v]-sc)*(n-m-siz[v]+sc))*w;
                 pp[c] = max( pp[c], tv );
             }
         }
     }
     dp[u][] = pp[];
     for( int c=; c<=maxc; c++ )
         dp[u][c] = max( pp[c], pp[c-] );
 } 

 int main() {
     scanf( "%d%d", &n, &m );
     if( n-m<m ) m=n-m;
     for( int i=,u,v,w; i<n; i++ ) {
         scanf( "%d%d%d", &u, &v, &w );
         adde( u, v, w );
         adde( v, u, w );
     }
     dfs(,);
     printf( "%lld\n", dp[][m] );
 }