poj - 3764

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的不行...加上之前瞎加的奇技淫巧的优化297ms跑进了前5

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define mp make_pair
 29 #define clr(a, x) memset(a, x, sizeof(a))
 30 
 31 const double pi = acos(-1.0);
 32 const int INF = 0x3f3f3f3f;
 33 const int MOD = 1e9 + 7;
 34 const double EPS = 1e-9;
 35 
 36 /*
 37 #include <ext/pb_ds/assoc_container.hpp>
 38 #include <ext/pb_ds/tree_policy.hpp>
 39 
 40 using namespace __gnu_pbds;
 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
 42 */
 43 
 44 const int MaxN = 100000;
 45 int n, p[MaxN + 15];
 46 struct Edge {
 47     int v, w;
 48     Edge () {}
 49     Edge (int v, int w) : v(v), w(w) {}
 50 } e[2 * MaxN + 15];
 51 
 52 int cnt, head[MaxN + 15], nex[MaxN + 15];
 53 void add_edge(int u, int v, int w) {
 54     e[++cnt] = Edge(v, w); nex[cnt] = head[u]; head[u] = cnt;
 55     e[++cnt] = Edge(u, w); nex[cnt] = head[v]; head[v] = cnt;
 56 }
 57 void dfs(int x, int pa) {
 58     for (int i = head[x]; i; i = nex[i]) {
 59         int y = e[i].v;
 60         if (y == pa) continue;
 61         p[y] = p[x] ^ e[i].w;
 62         dfs(y, x);
 63     }
 64 }
 65 int trie[2000015][2], tot;
 66 void add(int x) {
 67     int p = 1;
 68     for (int i = 30; ~i; --i) {
 69         int f = x >> i & 1;
 70         if (!trie[p][f]) {
 71             trie[p][f] = ++tot;
 72             trie[tot][0] = trie[tot][1] = 0;
 73         }
 74         p = trie[p][f];
 75     }
 76 }
 77 int query(int x) {
 78     int p = 1, y = 0;
 79     for (int i = 30; ~i; --i) {
 80         int f = x >> i & 1;
 81         if (trie[p][!f]) {
 82             y = y << 1 | !f;
 83             p = trie[p][!f];
 84         } else {
 85             y = y << 1 | f;
 86             p = trie[p][f];
 87         }
 88     }
 89     return x ^ y;
 90 }
 91 int getint() {
 92     int res = 0; char c = getchar();
 93     while (!isdigit(c)) c = getchar();
 94     while (isdigit(c)) {
 95         res = res * 10 + c - '0';
 96         c = getchar();
 97     }
 98     return res;
 99 }
100 int main() {
101     while (~scanf("%d", &n)) {
102         clr(head, 0), cnt = 0;
103         for (int i = 1; i < n; ++i) {
104             int u, v, w;
105             u = getint();
106             v = getint();
107             w = getint();
108             //scanf("%d%d%d", &u, &v, &w);
109             add_edge(u, v, w);
110         }
111         dfs(0, -1);
112         tot = 1; trie[tot][0] = trie[tot][1] = 0;
113         int ans = 0;
114         for (int i = 0; i < n; ++i) {
115             add(p[i]);
116             int res = query(p[i]);
117             ans = max(ans, res);
118         }
119         printf("%d\n", ans);
120     }
121     return 0;
122 }

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