BZOJ1954 Pku3764 The xor-longest Path

"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。

/**************************************************************
    Problem: 1954
    User: rausen
    Language: C++
    Result: Accepted
    Time:360 ms
    Memory:27368 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
const int N = 100005;
 
struct edge {
    int next, to, v;
    edge() {}
    edge(int _n, int _t, int _v) : next(_n), to(_t), v(_v) {}
} e[N << 1];
 
int first[N], tot;
 
struct trie_node {
    int son[2];
} t[3000005];
 
int cnt_trie = 1, root = 1;
 
int n, ans;
int v[N], a[40];
 
inline int read() {
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || '9' < ch) {
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
void Add_Edges(int x, int y, int z) {
    e[++tot] = edge(first[x], y, z), first[x] = tot;
    e[++tot] = edge(first[y], x, z), first[y] = tot;
}
 
void dfs(int p, int fa) {
    int x, y;
    for (x = first[p]; x; x = e[x].next)
        if ((y = e[x].to) != fa) {
            v[y] = v[p] ^ e[x].v;
            dfs(y, p);
        }
}
 
void insert(int x) {
    int now, i, cnt_a = 0;
    for (i = 0; i <= 30; ++i)
        a[i] = x & 1, x >>= 1;
    reverse(a, a + 31);
    now = root;
    for (i = 0; i <= 30; ++i) {
        if (!t[now].son[a[i]]) t[now].son[a[i]] = ++cnt_trie;
        now = t[now].son[a[i]];
    }
}
 
int query(int x) {
    int now, i, cnt_a = 0, res = 0;
    for (i = 0; i <= 30; ++i)
        a[i] = x & 1, x >>= 1;
    reverse(a, a + 31);
    now = root;
    for (i = 0; i <= 30; ++i) {
        if (t[now].son[!a[i]]) now = t[now].son[!a[i]], res += (1 << 30 - i);
        else now = t[now].son[a[i]];
    }
    return res;
}
 
int main() {
    int i, x, y, z; 
    n = read();
    for (i = 1; i < n; ++i) {
        x = read(), y = read(), z = read();
        Add_Edges(x, y, z);
    }
    dfs(1, 0);
    for (i = 1; i <= n; ++i)
        insert(v[i]);
    for (i = 1; i <= n; ++i)
        ans = max(ans, query(v[i]));
    printf("%d
", ans);
    return 0;
}