poj3764 The XOR Longest Path【dfs】【Trie树】

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10038		Accepted: 2040

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 uand 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)

Source

题意：

给一棵带边权的树，想找得到两个点，他们的路径上的权值异或最小。

思路：

首先我们任意找一个作为根，可以用dfs求出其他节点到根的路径的异或，记为xordis

那么对于树上的任意两个节点i, j，i到j的路径的异或和应该是xordis[i] ^ xordis[j]

因为i到j的路径，相当于i到根，根到j，其中重叠的部分，他们的异或值正好是0

因此这道题就变成了找两点异或值最小，https://www.cnblogs.com/wyboooo/p/9824293.html 和这道题就差不多了

最后还需要注意，search找到的最大值是除根以外的，还需要和xordis比较一下，取较大值。

 #include <iostream>
 #include <set>
 #include <cmath>
 #include <stdio.h>
 #include <cstring>
 #include <algorithm>
 #include <map>
 using namespace std;
 typedef long long LL;
 #define inf 0x7f7f7f7f

 int n;
 const int maxn = 1e5 + ;
 struct edge{
     int v, w;
     int nxt;
 }e[maxn * ];
 int head[maxn], tot = ;
 int xordis[maxn];
 int trie[maxn *  + ][], treetot = ;

 void addedge(int u, int v, int w)
 {
     e[tot].v = v;
     e[tot].w = w;
     e[tot].nxt = head[u];
     head[u] = tot++;
     e[tot].v = u;
     e[tot].w = w;
     e[tot].nxt = head[v];
     head[v] = tot++;
 }

 void dfs(int rt, int fa)
 {
     for(int i = head[rt]; i != -; i = e[i].nxt){
         int v = e[i].v;
         if(v == fa)continue;
         xordis[v] = xordis[rt] ^ e[i].w;
         dfs(v, rt);
     }
 }

 void init()
 {
     memset(head, -, sizeof(head));
     tot = ;
     memset(xordis, , sizeof(xordis));
     memset(trie, , sizeof(trie));
 }

 void insertt(int x)
 {
     int p = ;
     for(int i = ; i >= ; i--){
         int ch = x >> i & ;
         if(trie[p][ch] == ){
             trie[p][ch] = ++tot;
         }
         p = trie[p][ch];
     }
 }

 int searchh(int x)
 {
     int p = , ans = ;
     for(int i = ; i >= ; i--){
         int ch = x >> i & ;
         if(trie[p][ch ^ ]){
             p = trie[p][ch ^ ];
             ans |=  << i;
         }
         else{
             p = trie[p][ch];
         }
     }
     return ans;
 }

 int main()
 {
     while(scanf("%d", &n) != EOF){
         init();
         for(int i = ; i < n - ; i++){
             int u, v, w;
             scanf("%d%d%d", &u, &v, &w);
             addedge(u, v, w);
         }
         dfs(, -);

         /*for(int i = 0; i < n; i++){
             printf("%d\n", xordis[i]);
         }*/

         int ans = ;
         for(int i = ; i < n; i++){
             insertt(xordis[i]);
             //cout<<searchh(xordis[i])<<endl;
             ans = max(ans, searchh(xordis[i]));
             ans = max(ans, xordis[i]);
         }
         printf("%d\n", ans);
     }
     return ;
 }