卧薪尝胆,厚积薄发。
The xor-longest Path
Date: Mon Jul 16 20:58:05 CST 2018 In Category: NoCategory

Description:

求最长异或路径。
$1\le N \le10^5$

Solution:

先随便选一个点当根,然后求出每个点到根的异或距离并插到一棵 $trie$ 树中,枚举每个点查与它异或的最大值即可。

Code:


#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int n;
#define MAXN 100010
typedef unsigned int uint;
struct node
{
int tr[2];
}t[MAXN * 32];
int ptr = 0;
int newnode(){return ++ptr;}
int root;
void insert(uint k)
{
int cur = root;
for(int i = 31;i >= 0;--i)
{
if(t[cur].tr[(k >> i) & 1] == 0)t[cur].tr[(k >> i) & 1] = newnode();
cur = t[cur].tr[(k >> i) & 1];
}
return;
}
int query(uint k)
{
int cur = root;
uint res = 0;
for(int i = 31;i >= 0;--i)
{
if(t[cur].tr[!((k >> i) & 1)] != 0)
{
cur = t[cur].tr[!((k >> i) & 1)];
res += (1 << i);
}
else cur = t[cur].tr[(k >> i) & 1];
}
return res;
}
struct edge
{
int to,nxt;
uint v;
}e[MAXN << 1];
int edgenum = 0;
int lin[MAXN] = {0};
void add(int a,int b,uint c)
{
++edgenum;e[edgenum].to = b;e[edgenum].v = c;e[edgenum].nxt = lin[a];lin[a] = edgenum;
++edgenum;e[edgenum].to = a;e[edgenum].v = c;e[edgenum].nxt = lin[b];lin[b] = edgenum;
return;
}
bool v[MAXN];
uint val[MAXN];
void dfs(int k)
{
v[k] = true;
insert(val[k]);
for(int i = lin[k];i != 0;i = e[i].nxt)
{
if(v[e[i].to])continue;
val[e[i].to] = val[k] ^ e[i].v;
dfs(e[i].to);
}
return;
}
int main()
{
while(scanf("%d",&n) != EOF)
{
memset(t,0,sizeof(t));
ptr = 0;
memset(lin,0,sizeof(lin));
edgenum = 0;
int a,b;
uint c;
for(int i = 1;i < n;++i)
{
scanf("%d%d%d",&a,&b,&c);
++a;++b;
add(a,b,c);
}
memset(v,false,sizeof(v));
dfs(1);
int ans = 0;
for(int i = 1;i <= n;++i)
{
ans = max(ans,query(val[i]));
}
cout << ans << endl;
}
return 0;
}
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