"trie的经典应用" -- by hzwer
我们把每个点到根的xor值记下来，然后找出两个xor值最大的即可（因为(a ^ c) ^ (b ^ c) = a ^ b）
于是用trie把每个数的二进制位记下来，每次query的时候利用贪心，试图走到另一个儿子即可。
1 /************************************************************** 2 Problem: 1954 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:360 ms 7 Memory:27368 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 100005; 15 16 struct edge { 17 int next, to, v; 18 edge() {} 19 edge(int _n, int _t, int _v) : next(_n), to(_t), v(_v) {} 20 } e[N << 1]; 21 22 int first[N], tot; 23 24 struct trie_node { 25 int son[2]; 26 } t[3000005]; 27 28 int cnt_trie = 1, root = 1; 29 30 int n, ans; 31 int v[N], a[40]; 32 33 inline int read() { 34 int x = 0, sgn = 1; 35 char ch = getchar(); 36 while (ch < '0' || '9' < ch) { 37 if (ch == '-') sgn = -1; 38 ch = getchar(); 39 } 40 while ('0' <= ch && ch <= '9') { 41 x = x * 10 + ch - '0'; 42 ch = getchar(); 43 } 44 return sgn * x; 45 } 46 47 void Add_Edges(int x, int y, int z) { 48 e[++tot] = edge(first[x], y, z), first[x] = tot; 49 e[++tot] = edge(first[y], x, z), first[y] = tot; 50 } 51 52 void dfs(int p, int fa) { 53 int x, y; 54 for (x = first[p]; x; x = e[x].next) 55 if ((y = e[x].to) != fa) { 56 v[y] = v[p] ^ e[x].v; 57 dfs(y, p); 58 } 59 } 60 61 void insert(int x) { 62 int now, i, cnt_a = 0; 63 for (i = 0; i <= 30; ++i) 64 a[i] = x & 1, x >>= 1; 65 reverse(a, a + 31); 66 now = root; 67 for (i = 0; i <= 30; ++i) { 68 if (!t[now].son[a[i]]) t[now].son[a[i]] = ++cnt_trie; 69 now = t[now].son[a[i]]; 70 } 71 } 72 73 int query(int x) { 74 int now, i, cnt_a = 0, res = 0; 75 for (i = 0; i <= 30; ++i) 76 a[i] = x & 1, x >>= 1; 77 reverse(a, a + 31); 78 now = root; 79 for (i = 0; i <= 30; ++i) { 80 if (t[now].son[!a[i]]) now = t[now].son[!a[i]], res += (1 << 30 - i); 81 else now = t[now].son[a[i]]; 82 } 83 return res; 84 } 85 86 int main() { 87 int i, x, y, z; 88 n = read(); 89 for (i = 1; i < n; ++i) { 90 x = read(), y = read(), z = read(); 91 Add_Edges(x, y, z); 92 } 93 dfs(1, 0); 94 for (i = 1; i <= n; ++i) 95 insert(v[i]); 96 for (i = 1; i <= n; ++i) 97 ans = max(ans, query(v[i])); 98 printf("%d ", ans); 99 return 0; 100 }