POJ3764 The xor-longest Path

Posted 2020-08-23 SilverNebula

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6361		Accepted: 1378

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)

Source

 

trie树+贪心

DFS预处理出每个结点到根的路径的异或和。两点之间路径的异或和等于各自到根的路径的异或和的异或。

将所有的异或和转化成二进制串，建成trie树。

对于每个二进制串，在trie树上贪心选取使异或值最大的路径（尽量通往数值相反的结点），记录最优答案。

↑为了防止匹配到自身的一部分，每个二进制串都应该先查完再插入trie树。

 

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<vector>
 8 using namespace std;
 9 const int mxn=200010;
10 int read(){
11     int x=0,f=1;char ch=getchar();
12     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
13     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
14     return x*f;
15 }
16 struct edge{
17     int v,nxt,w;
18 }e[mxn<<1];
19 int hd[mxn],mct=0;
20 void add_edge(int u,int v,int w){
21     e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return;
22 }
23 int t[mxn*10][2],cnt=1;
24 int n,ans=0;
25 int f[mxn];
26 void insert(int x){
27     int now=1;
28     for(int i=31;i>=0;i--){
29         int v=(x>>i)&1;
30         if(!t[now][v])t[now][v]=++cnt;
31         now=t[now][v];
32     }
33     return;
34 }
35 void query(int x){
36     int now=1;
37     int res=0;
38     for(int i=31;i>=0;i--){
39         int v=(x>>i)&1;
40         if(t[now][v^1]){
41             v^=1;
42             res+=(1<<i);
43         }
44         now=t[now][v];
45     }
46     ans=max(res,ans);
47     return;
48 }
49 void DFS(int u,int fa,int num){
50     f[u]=num;
51     for(int i=hd[u];i;i=e[i].nxt){
52         int v=e[i].v;
53         if(v==fa)continue;
54         DFS(v,u,num^e[i].w);
55     }
56     return;
57 }
58 void init(){
59     memset(hd,0,sizeof hd);
60     memset(t,0,sizeof t);
61 //    memset(f,0,sizeof f);
62     mct=0;cnt=1;ans=0;
63     return;
64 }
65 int main(){
66     while(scanf("%d",&n)!=EOF){
67         init();
68         int i,j,u,v,w;
69         for(i=1;i<n;i++){
70             u=read();v=read();w=read();
71             u++;v++;
72             add_edge(u,v,w);
73             add_edge(v,u,w);
74         }
75         DFS(1,0,0);
76         for(i=1;i<=n;i++){
77 //            printf("f[%d]:%d\n",i,f[i]);
78             query(f[i]);
79             insert(f[i]);
80         }
81         printf("%d\n",ans);
82     }
83     return 0;
84 }