Tsinsen-1486:树【Trie树 + 点分治】

暴力部分:

　　这个题一开始的想法是 n^2 枚举两个点，然后logn维护LCA，在倍增的同时维护异或值和 k 的个数。

　　s_z_l老爷指导了新的思路，既然这个树只有n^2个LCA，那么枚举LCA，同时向下深搜即可。

标算:

　　首先点分治，尽力保证树的平衡，然后按照Trie树的性质，贪心，至于k，我们可以把每个节点的值置为like值的最大值，然后在走左右儿子的时候判断一下即可。

　　点分治时 !vis[E[i].v] && E[i].v != fa 一定不要忘

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define travel(x) for (int i = G[x]; i; i = E[i].nx)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first
 #define yy second
 using namespace std;
 typedef long long i64;
 typedef pair<int, int> pii;
 //********************************
 const int maxn = ;
 struct Ed {
     int u, v, nx; Ed() {}
     Ed(int _u, int _v, int _nx) :
         u(_u), v(_v), nx(_nx) {}
 } E[maxn << ];
 int G[maxn], cnt_e;
 void addedge(int u, int v) {
     E[++cnt_e] = Ed(u, v, G[u]);
     G[u] = cnt_e;
 }
 int vis[maxn];
 pii sta[maxn]; int top;
 int rt;
 int f[maxn], size[maxn], sum;
 void getrt(int x, int fa)
 {
     f[x] = , size[x] = ;
     travel(x) {
         if (!vis[E[i].v] && E[i].v != fa) {
             getrt(E[i].v, x);
             size[x] += size[E[i].v];
             f[x] = max(f[x], size[E[i].v]);
         }
     }
     f[x] = max(f[x], sum - size[x]);
     if (f[x] < f[rt]) rt = x;
 }
 int read() {
     int l = , s(); char ch = getchar();
     while (ch < '' || ch > '') { if (ch == '-') l = -; ch = getchar(); }
     while (ch >= ''&& ch <= '') { s = (s << ) + (s << ) + ch - ''; ch = getchar(); }
     return l * s;
 }
 int trie[][], tab[], rootr;
 int ntot;
 int n, K;
 int c[maxn], w[maxn];
 int query(int co, int k) {
     int ret();
     int p = rootr;
     if (tab[p] + k < K) return -;
     drep(i, , ) {
         int id = co >> i & ;
         if (trie[p][id ^ ] && tab[trie[p][id ^ ]] + k >= K) ret |=  << i, p = trie[p][id ^ ];
         else if (!trie[p][id] || tab[trie[p][id]] + k < K) { ret = -; break; }
         else p = trie[p][id];
     }
     return ret;
 }
 void insrt(int co, int k) {
     int p = rootr;
     drep(i, , ) {
         tab[p] = max(tab[p], k);
         int id = co >> i & ;
         if (!trie[p][id]) trie[p][id] = ++ntot;
         p = trie[p][id];
     }
     tab[p] = max(tab[p], k);
 }
 int ans = -;
 void dfs_query(int x, int fa, int co, int k) {
     sta[++top] = mp(co, k);
     ans = max(ans, query(co, k));
     for (int i = G[x]; i; i = E[i].nx) if (!vis[E[i].v] && E[i].v != fa)
         dfs_query(E[i].v, x, co ^ w[E[i].v], k + c[E[i].v]);
 }
 /*
    void dfs_insrt(int x, int fa, int co, int k) {
    insrt(rootr, co, k, 30);
    for (int i = G[x]; i; i = E[i].nx) if (!vis[E[i].v] && E[i].v != fa)
    dfs_insrt(E[i].v, x, co ^ w[E[i].v], k + c[E[i].v]);
    }
  */
 void solve(int x) {
     vis[x] = ;
     rep(i, , ntot) trie[i][] = trie[i][] = , tab[i] = ;
     rootr = ;
     ntot = ;
     if (c[x] >= K) ans = max(ans, w[x]);
     insrt(w[x], c[x]);
     travel(x) {
         if (vis[E[i].v]) continue;
         top = ;
         dfs_query(E[i].v, x, w[E[i].v], c[E[i].v]);
         rep(j, , top) {
             sta[j].xx ^= w[x], sta[j].yy += c[x];
             insrt(sta[j].xx, sta[j].yy);
         }
     }
     int tmp = sum;
     for (int i = G[x]; i; i = E[i].nx) {
         if (!vis[E[i].v]) {
             rt = ; sum = size[E[i].v] > size[x] ? tmp - size[x] : size[E[i].v];
             getrt(E[i].v, );
             solve(rt);
         }
     }
 }
 int main() {
     n = read(), K = read();
     rep(i, , n) c[i] = read();
     rep(i, , n) w[i] = read();
     REP(i, , n) {
         int x, y; x = read(), y = read();
         addedge(x, y), addedge(y, x);
     }
     rt = , sum = n, f[] = n + ;
     getrt(, );
     solve(rt);
     printf("%d\n", ans);
     return ;
 }