poj

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8273		Accepted: 1720

Description

In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p:

⊕ is the xor operator.

We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path? 　

Input

The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.

Output

For each test case output the xor-length of the xor-longest path.

Sample Input

4
0 1 3
1 2 4
1 3 6

Sample Output

7

Hint

The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)

思路： 每个点存一下到根节点路径的异或和，这些值两两异或的最大值就是答案。

　　POJ不开O2，vector慢的不行，一直以为字典树写搓了TLE到绝望，试了三种不同的字典树写法，结果换成链式前向星快得飞起...加上之前瞎加的奇技淫巧的优化297ms跑进了前5

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

const int MaxN = 100000;
int n, p[MaxN + 15];
struct Edge {
    int v, w;
    Edge () {}
    Edge (int v, int w) : v(v), w(w) {}
} e[2 * MaxN + 15];

int cnt, head[MaxN + 15], nex[MaxN + 15];
void add_edge(int u, int v, int w) {
    e[++cnt] = Edge(v, w); nex[cnt] = head[u]; head[u] = cnt;
    e[++cnt] = Edge(u, w); nex[cnt] = head[v]; head[v] = cnt;
}
void dfs(int x, int pa) {
    for (int i = head[x]; i; i = nex[i]) {
        int y = e[i].v;
        if (y == pa) continue;
        p[y] = p[x] ^ e[i].w;
        dfs(y, x);
    }
}
int trie[2000015][2], tot;
void add(int x) {
    int p = 1;
    for (int i = 30; ~i; --i) {
        int f = x >> i & 1;
        if (!trie[p][f]) {
            trie[p][f] = ++tot;
            trie[tot][0] = trie[tot][1] = 0;
        }
        p = trie[p][f];
    }
}
int query(int x) {
    int p = 1, y = 0;
    for (int i = 30; ~i; --i) {
        int f = x >> i & 1;
        if (trie[p][!f]) {
            y = y << 1 | !f;
            p = trie[p][!f];
        } else {
            y = y << 1 | f;
            p = trie[p][f];
        }
    }
    return x ^ y;
}
int getint() {
    int res = 0; char c = getchar();
    while (!isdigit(c)) c = getchar();
    while (isdigit(c)) {
        res = res * 10 + c - '0';
        c = getchar();
    }
    return res;
}
int main() {
    while (~scanf("%d", &n)) {
        clr(head, 0), cnt = 0;
        for (int i = 1; i < n; ++i) {
            int u, v, w;
            u = getint();
            v = getint();
            w = getint();
            //scanf("%d%d%d", &u, &v, &w);
            add_edge(u, v, w);
        }
        dfs(0, -1);
        tot = 1; trie[tot][0] = trie[tot][1] = 0;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            add(p[i]);
            int res = query(p[i]);
            ans = max(ans, res);
        }
        printf("%d
", ans);
    }
    return 0;
}