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。
1 #include <iostream>
2 #include <set>
3 #include <cmath>
4 #include <stdio.h>
5 #include <cstring>
6 #include <algorithm>
7 #include <map>
8 using namespace std;
9 typedef long long LL;
10 #define inf 0x7f7f7f7f
11
12 int n;
13 const int maxn = 1e5 + 5;
14 struct edge{
15 int v, w;
16 int nxt;
17 }e[maxn * 2];
18 int head[maxn], tot = 0;
19 int xordis[maxn];
20 int trie[maxn * 32 + 5][3], treetot = 1;
21
22 void addedge(int u, int v, int w)
23 {
24 e[tot].v = v;
25 e[tot].w = w;
26 e[tot].nxt = head[u];
27 head[u] = tot++;
28 e[tot].v = u;
29 e[tot].w = w;
30 e[tot].nxt = head[v];
31 head[v] = tot++;
32 }
33
34 void dfs(int rt, int fa)
35 {
36 for(int i = head[rt]; i != -1; i = e[i].nxt){
37 int v = e[i].v;
38 if(v == fa)continue;
39 xordis[v] = xordis[rt] ^ e[i].w;
40 dfs(v, rt);
41 }
42 }
43
44 void init()
45 {
46 memset(head, -1, sizeof(head));
47 tot = 0;
48 memset(xordis, 0, sizeof(xordis));
49 memset(trie, 0, sizeof(trie));
50 }
51
52 void insertt(int x)
53 {
54 int p = 1;
55 for(int i = 30; i >= 0; i--){
56 int ch = x >> i & 1;
57 if(trie[p][ch] == 0){
58 trie[p][ch] = ++tot;
59 }
60 p = trie[p][ch];
61 }
62 }
63
64 int searchh(int x)
65 {
66 int p = 1, ans = 0;
67 for(int i = 30; i >= 0; i--){
68 int ch = x >> i & 1;
69 if(trie[p][ch ^ 1]){
70 p = trie[p][ch ^ 1];
71 ans |= 1 << i;
72 }
73 else{
74 p = trie[p][ch];
75 }
76 }
77 return ans;
78 }
79
80 int main()
81 {
82 while(scanf("%d", &n) != EOF){
83 init();
84 for(int i = 0; i < n - 1; i++){
85 int u, v, w;
86 scanf("%d%d%d", &u, &v, &w);
87 addedge(u, v, w);
88 }
89 dfs(0, -1);
90
91 /*for(int i = 0; i < n; i++){
92 printf("%d\n", xordis[i]);
93 }*/
94
95 int ans = 0;
96 for(int i = 1; i < n; i++){
97 insertt(xordis[i]);
98 //cout<<searchh(xordis[i])<<endl;
99 ans = max(ans, searchh(xordis[i]));
100 ans = max(ans, xordis[i]);
101 }
102 printf("%d\n", ans);
103 }
104 return 0;
105 }
原文地址：https://www.cnblogs.com/wyboooo/p/9824427.html