【Trie树 + 点分治】Tsinsen1486:树

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\n", ans);
126     return 0;
127 }