LuoGuP4551最长异或路径

LuoGuP4551最长异或路径
(01Trie)裸题,懒得写(solution)了,直接贴代码吧,好懒啊yyy.
(Code:)

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
#define MEM(x,y) memset ( x , y , sizeof ( x ) )
#define rep(i,a,b) for (int i = a ; i <= b ; ++ i)
#define per(i,a,b) for (int i = a ; i >= b ; -- i)
#define pii pair < int , int >
#define X first
#define Y second
#define rint read<int>
#define int long long
#define pb push_back

using std::set ;
using std::pair ;
using std::max ;
using std::min ;
using std::priority_queue ;
using std::vector ;
using std::swap ;
using std::sort ;
using std::unique ;
using std::greater ;

template < class T >
    inline T read () {
        T x = 0 , f = 1 ; char ch = getchar () ;
        while ( ch < '0' || ch > '9' ) {
            if ( ch == '-' ) f = - 1 ;
            ch = getchar () ;
        }
        while ( ch >= '0' && ch <= '9' ) {
            x = ( x << 3 ) + ( x << 1 ) + ( ch - 48 ) ;
            ch = getchar () ;
       }
   return f * x ;
}

const int N = 1e5 + 100 ;
const int maxsize = 1 << 30 ;

struct edge { int to , next , data ; } e[(N<<2)] ;

int dis[N] , n , head[N] , tot , ch[N*100][2] , cnt , sum[N*100] , ans ;

inline void dfs (int cur , int anc , int dist) {
    dis[cur] = dist ;
    for (int i = head[cur] ; i ; i = e[i].next) {
        int k = e[i].to ;
        if ( k == anc ) continue ;
        dfs ( k , cur , dist ^ e[i].data ) ;
    }
    return ;
}

inline void insert (int cur) {
    int now = 0 ;
    for (int i = maxsize ; i ; i >>= 1) {
        bool son = ( ( i & cur ) != 0 ) ? true : false ;
        if ( ! ch[now][son] ) ch[now][son] = ++ cnt ;
        now = ch[now][son] ;
    }
    ++ sum[now] ; return ;
}

inline int greedy (int cur) {
    int now = 0 , res = 0 ;
    for (int i = maxsize ; i ; i >>= 1) {
        bool son = ( ( i & cur ) != 0 ) ? false : true ;
        if ( ! ch[now][son] ) son ^= 1 ;
        res = ( res << 1 | son ) ;
        now = ch[now][son] ;
    }
    return res ;
}

inline void build (int u , int v , int w) {
    e[++tot].next = head[u] ; e[tot].to = v ;
    e[tot].data = w ; head[u] = tot ; return;
}

signed main (int argc , char * argv[] ) {
    n = rint () ;
    rep ( i , 2 , n ) {
        int u = rint () , v = rint () , w = rint () ;
        build ( u , v , w ) ; build ( v , u , w ) ;
    }
    dfs ( 1 , 0 , 0 ) ;
    rep ( i , 1 , n ) insert ( dis[i] ) ;
    rep ( i , 1 , n ) ans = max ( ans , dis[i] ^ greedy ( dis[i] ) ) ;
    printf ("%lld
" , ans ) ;
    system ("pause") ; return 0 ;
}