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
1 #include <bits/stdc++.h> 2 #define rep(i, a, b) for (int i = a; i <= b; i++) 3 #define REP(i, a, b) for (int i = a; i < b; i++) 4 #define drep(i, a, b) for (int i = a; i >= b; i--) 5 #define travel(x) for (int i = G[x]; i; i = E[i].nx) 6 #define mp make_pair 7 #define pb push_back 8 #define clr(x) memset(x, 0, sizeof(x)) 9 #define xx first 10 #define yy second 11 using namespace std; 12 typedef long long i64; 13 typedef pair<int, int> pii; 14 //******************************** 15 const int maxn = 100005; 16 struct Ed { 17 int u, v, nx; Ed() {} 18 Ed(int _u, int _v, int _nx) : 19 u(_u), v(_v), nx(_nx) {} 20 } E[maxn << 1]; 21 int G[maxn], cnt_e; 22 void addedge(int u, int v) { 23 E[++cnt_e] = Ed(u, v, G[u]); 24 G[u] = cnt_e; 25 } 26 int vis[maxn]; 27 pii sta[maxn]; int top; 28 int rt; 29 int f[maxn], size[maxn], sum; 30 void getrt(int x, int fa) 31 { 32 f[x] = 0, size[x] = 1; 33 travel(x) { 34 if (!vis[E[i].v] && E[i].v != fa) { 35 getrt(E[i].v, x); 36 size[x] += size[E[i].v]; 37 f[x] = max(f[x], size[E[i].v]); 38 } 39 } 40 f[x] = max(f[x], sum - size[x]); 41 if (f[x] < f[rt]) rt = x; 42 } 43 int read() { 44 int l = 1, s(0); char ch = getchar(); 45 while (ch < '0' || ch > '9') { if (ch == '-') l = -1; ch = getchar(); } 46 while (ch >= '0'&& ch <= '9') { s = (s << 1) + (s << 3) + ch - '0'; ch = getchar(); } 47 return l * s; 48 } 49 int trie[3000005][2], tab[3000005], rootr; 50 int ntot; 51 int n, K; 52 int c[maxn], w[maxn]; 53 int query(int co, int k) { 54 int ret(0); 55 int p = rootr; 56 if (tab[p] + k < K) return -1; 57 drep(i, 30, 0) { 58 int id = co >> i & 1; 59 if (trie[p][id ^ 1] && tab[trie[p][id ^ 1]] + k >= K) ret |= 1 << i, p = trie[p][id ^ 1]; 60 else if (!trie[p][id] || tab[trie[p][id]] + k < K) { ret = -1; break; } 61 else p = trie[p][id]; 62 } 63 return ret; 64 } 65 void insrt(int co, int k) { 66 int p = rootr; 67 drep(i, 30, 0) { 68 tab[p] = max(tab[p], k); 69 int id = co >> i & 1; 70 if (!trie[p][id]) trie[p][id] = ++ntot; 71 p = trie[p][id]; 72 } 73 tab[p] = max(tab[p], k); 74 } 75 int ans = -1; 76 void dfs_query(int x, int fa, int co, int k) { 77 sta[++top] = mp(co, k); 78 ans = max(ans, query(co, k)); 79 for (int i = G[x]; i; i = E[i].nx) if (!vis[E[i].v] && E[i].v != fa) 80 dfs_query(E[i].v, x, co ^ w[E[i].v], k + c[E[i].v]); 81 } 82 /* 83 void dfs_insrt(int x, int fa, int co, int k) { 84 insrt(rootr, co, k, 30); 85 for (int i = G[x]; i; i = E[i].nx) if (!vis[E[i].v] && E[i].v != fa) 86 dfs_insrt(E[i].v, x, co ^ w[E[i].v], k + c[E[i].v]); 87 } 88 */ 89 void solve(int x) { 90 vis[x] = 1; 91 rep(i, 0, ntot) trie[i][0] = trie[i][1] = 0, tab[i] = 0; 92 rootr = 0; 93 ntot = 0; 94 if (c[x] >= K) ans = max(ans, w[x]); 95 insrt(w[x], c[x]); 96 travel(x) { 97 if (vis[E[i].v]) continue; 98 top = 0; 99 dfs_query(E[i].v, x, w[E[i].v], c[E[i].v]); 100 rep(j, 1, top) { 101 sta[j].xx ^= w[x], sta[j].yy += c[x]; 102 insrt(sta[j].xx, sta[j].yy); 103 } 104 } 105 int tmp = sum; 106 for (int i = G[x]; i; i = E[i].nx) { 107 if (!vis[E[i].v]) { 108 rt = 0; sum = size[E[i].v] > size[x] ? tmp - size[x] : size[E[i].v]; 109 getrt(E[i].v, 0); 110 solve(rt); 111 } 112 } 113 } 114 int main() { 115 n = read(), K = read(); 116 rep(i, 1, n) c[i] = read(); 117 rep(i, 1, n) w[i] = read(); 118 REP(i, 1, n) { 119 int x, y; x = read(), y = read(); 120 addedge(x, y), addedge(y, x); 121 } 122 rt = 0, sum = n, f[0] = n + 1; 123 getrt(1, 0); 124 solve(rt); 125 printf("%d ", ans); 126 return 0; 127 }