poj3764字典树路径最大异或和

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6853		Accepted: 1464

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)

先求出来到节点1的，然后字典树从高位到低位。异或和的性质以及贪心？。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e5+;
int head[N],tot,sz,val[N];
struct node
{
    int v,next,w;
} e[N<<];
struct Trie
{
    int next[];
    void init()
    {
        memset(next,,sizeof(next));
    }
} Ti[N*];
void add(int u,int v,int w)
{
    e[tot].v=v;
    e[tot].next=head[u];
    e[tot].w=w;
    head[u]=tot++;
    e[tot] .v=u;
    e[tot].next=head[v];
    e[tot].w=w;
    head[v]=tot++;
}
void Add(int x)
{
    int u=,ind;
    for(int i=; i>=; --i)
    {
        if((<<i)&x) ind=;
        else ind=;
        if(!Ti[u].next[ind])
        {
            Ti[u].next[ind]=++sz;
            Ti[sz].init();
        }
        u=Ti[u].next[ind];
    }
}
int get(int x)
{
    int u=,num=,ind;
    for(int i=; i>=; --i)
    {
        if((<<i)&x) ind=;
        else ind=;
        if(Ti[u].next[ind])
        {
            u=Ti[u].next[ind];
            num|=(<<i);
        }
        else u=Ti[u].next[!ind];
    }
    return num;
}
void dfs(int x,int v,int fa)
{
    val[x]=v;
    for(int i=head[x]; i+; i=e[i].next)
    {
        int to=e[i].v;
        if(to==fa) continue;
        dfs(to,e[i].w^v,x);
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        tot=sz=;
        int x,y,z;
        memset(head,-,sizeof(head));
        for(int i=; i<n; ++i)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x+,y+,z);
        }
        Ti[].init();  //必须初始化
        dfs(,,);
        int maxx=;
        for(int i=; i<=n; ++i)
        {
            maxx=max(maxx,get(val[i]));
            Add(val[i]);
        }
        printf("%d\n",maxx);
    }
}