HDU 6060 RXD and dividing

题意：
给出一个具有n个顶点的树， 将其中的2~n分成k部分， 求1号顶点到分成k块后各个顶点的最大权值和。
思路：
将2~n的结点编号{1~k}，每条边的权乘以其下子树的大小与k的较小值。
代码：

# include <bits/stdc++.h>using namespace std;typedef long long LL;const int INF = 0x3f3f3f3f;const int Q = 1e9+7;const int N = 1e6+10;int n, k, tot;int head[N], f[N], si[N], w[N];struct Edge {    int fr, to, co, ne;} E[N * 2];void add (int u, int v, int w){    E[tot].fr = u;    E[tot].to = v;    E[tot].co = w;    E[tot].ne = head[u];    head[u] = tot++;}void dfs (int u, int fa){    si[u] = 1;    for (int i=head[u]; i!=-1; i=E[i].ne) {        int to = E[i].to;        if (to == fa) continue;        w[to] = E[i].co;        dfs(to, u);        si[u] += si[to];    }}void solve (){    LL ans = 0;    tot = 0;    memset (head, -1, sizeof (head));    memset (si, 0, sizeof (si));    for (int i=0; i<n-1; ++i) {        int u, v, w;        scanf ("%d%d%d", &u, &v, &w);        add(u, v, w);        add(v, u, w);    }    dfs (1, -1);    for (int i=2; i<=n; ++i) {        ans += (LL)w[i] * min(si[i], k);    }    printf ("%lld\n", ans);}int main (){    while (scanf ("%d%d", &n, &k) != EOF) {        solve ();    }    return 0;}