BZOJ 2599: [IOI2011]Race

点分治，定权值，求另一关键字最小

不满足前缀加减性

可以按序遍历，用一数组$t[] 来维护路径为i的最小边数$

再对于一个直系儿子对应的子树，先算距离求答案再更新$t数组，这样就不会重复$

#include <bits/stdc++.h>
using namespace std;

#define N 200010
#define M 1000010
#define INF 0x3f3f3f3f
int n, k;
struct Graph
{
    struct node
    {
        int to, nx, w;
        node() {}
        node(int to, int nx, int w) : to(to), nx(nx), w(w) {}
    }a[N << 1];
    int head[N], pos;
    void init()
    {
        memset(head, 0, sizeof head);
        pos = 0;
    }
    void add(int u, int v, int w)
    {
        a[++pos] = node(v, head[u], w); head[u] = pos;
        a[++pos] = node(u, head[v], w); head[v] = pos;
    }
}G;
#define erp(u) for (int it = G.head[u], v = G.a[it].to, w = G.a[it].w; it; it = G.a[it].nx, v = G.a[it].to, w = G.a[it].w)

int vis[N];
int sum, root, sze[N], f[N];
void getroot(int u, int fa)
{
    f[u] = 0, sze[u] = 1;
    erp(u) if (v != fa && !vis[v])
    {
        getroot(v, u);
        sze[u] += sze[v];
        f[u] = max(f[u], sze[v]);
    }
    f[u] = max(f[u], sum - sze[u]);
    if (f[u] < f[root]) root = u;
}

int dist[N], deep[N], t[M], res;
void getdeep(int u, int fa)
{
    if (dist[u] <= k) res = min(res, deep[u] + t[k - dist[u]]);
    erp(u) if (v != fa && !vis[v])
    {
        dist[v] = dist[u] + w;
        deep[v] = deep[u] + 1;
        getdeep(v, u);   
    }
}

void add(int u, int fa, int flag)
{
    if (dist[u] <= k)
    {
        if (flag) t[dist[u]] = min(t[dist[u]], deep[u]);
        else t[dist[u]] = INF; 
    }
    erp(u) if (v != fa && !vis[v])
        add(v, u, flag);
}

void solve(int u)
{
    vis[u] = 1; t[0] = 0;
    erp(u) if (!vis[v])
    {
        deep[v] = 1, dist[v] = w;
        getdeep(v, u); add(v, u, 1);
    }
    erp(u) if (!vis[v]) add(v, u, 0);
    erp(u) if (!vis[v])
    {
        sum = f[0] = sze[v]; root = 0;
        getroot(v, 0);
        solve(root);
    }
}

void Run()
{
    while (scanf("%d%d", &n, &k) != EOF)
    {
        G.init(); res = INF;
        memset(t, 0x3f, sizeof t);
        for (int i = 1, u, v, w; i < n; ++i)
        {
            scanf("%d%d%d", &u, &v, &w);
            ++u, ++v;
            G.add(u, v, w);
        }
        sum = f[0] = n; root = 0;
        getroot(1, 0);
        solve(root);
        printf("%d
", res == INF ? -1 : res); 
    }
} 

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif 

    Run();
    return 0;
}