POJ 3764 The xor-longest Path

Posted 2020-12-31 traveller-ly

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

题目说明：给定一个N个节点的数，树上每条边都有1个权值。从树中选择两个点x和y，把从x到y的路径上的所有的边权XOR（异或）起来，求最大的结果。

做法：比较模板的一道字典树题，设D[x]表示根到x上所有边权的XOR值，那么x到y的XOR值显然为D[x]xor D[y]。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #define rep(i,a,b) for(int i=a;i<=b;i++)
 5 #define fx(i,x) for(int i=ls[x];i;i=e[i].next)
 6 #define N 100009
 7 #define LL long long
 8 #define fill(a,b) memset(a,b,sizeof(a))
 9 using namespace std;
10 int ls[N],n,tot,t[N*60][2],cnt,ans;
11 int d[N];
12 struct edge{
13     int to,next,w;
14 }e[N*2];
15 
16 inline int max(int a,int b) {return a>b?a:b;}
17 
18 inline void Add(int x,int y,int z){
19     e[++tot].to=y;
20     e[tot].next=ls[x];
21     e[tot].w=z;
22     ls[x]=tot;
23 }
24 
25 inline int Read(){
26     int s=0;
27     char ch=getchar();
28     for(;ch<‘0‘||ch>‘9‘;ch=getchar());
29     for(;ch>=‘0‘&&ch<=‘9‘;s=(s<<3)+(s<<1)+(ch^48),ch=getchar());
30     return s;
31 }
32 
33 void Init(){
34     fill(d,0);
35     fill(ls,0);
36     fill(t,0);
37     fill(e,0);
38     tot=0;
39     ans=0;
40     cnt=0;
41     rep(i,1,n-1){
42         int u,v,w;
43         u=Read(),v=Read(),w=Read();
44         Add(u,v,w),Add(v,u,w);
45     }
46         
47 }
48 
49 inline void Dfs(int x,int fa){
50     fx(i,x){
51         if (e[i].to==fa) continue;
52         d[e[i].to]=d[x]^e[i].w;
53         Dfs(e[i].to,x);
54     }
55 }
56 
57 inline void Insert(int x){
58     int root=0,ch;
59     for(int i=30;i>=0;i--){
60         if (x&(1<<i)) ch=1; else ch=0;
61         if (!t[root][ch])    t[root][ch]=++cnt;
62         root=t[root][ch];
63     }
64 }
65 
66 inline int Search(int x){
67     int root=0,sum=0,ch;
68     for(int i=30;i>=0;i--){
69         if(x&(1<<i)) ch=0; else ch=1;
70         if(t[root][ch]){
71             sum|=(1<<i);
72             root=t[root][ch];
73         }else root=t[root][!ch];    
74     }
75     return sum;
76 }
77 
78 void Work(){
79     rep(i,0,n) Insert(d[i]);
80     rep(i,0,n) ans=max(ans,Search(d[i]));
81 }
82 
83 int main(){
84     while(~scanf("%d",&n)){
85         Init();
86         Dfs(1,n+2);
87         Work();
88         printf("%d
",ans);
89     }
90 }

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