「BZOJ1954」Pku3764 The xor – longest Path

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给定一棵n个点的带权树，求树上最长的异或和路径
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
1 2 3
2 3 4
2 4 6
Sample Output
7

HINT

The xor-longest path is 1->2->3, which has length 7 (=3 ⊕ 4)
注意：结点下标从1开始到N….
求出根到每个结点的边权异或和
问题转换为任选两点，异或和最大
trie的经典？应用
在trie上从插入所有权值，再查询每个权值。。。（从高位到低位）
查询x的时候，每一位尽量走与x该位不同的结点

#include<bits/stdc++.h>
using namespace std;
#define node edge[i].to
int trie[4000000][2],n,dis[100001],head[100001],cnt,tot;
struct tu{
    int to,pre,f;
}edge[200000];
inline void insert(int x)
{
    int p=0;
    for(int i=31;i>=0;i--)
	{
        bool f=((x&(1<<i))!=0);
        if(!trie[p][f])trie[p][f]=++tot;
        p=trie[p][f];
    }
}
inline void add(int x,int y,int z)
{
    edge[++cnt].f=z;
    edge[cnt].to=y;
    edge[cnt].pre=head[x];
    head[x]=cnt;
}
inline void dfs(int x,int fa)
{
    insert(dis[x]);//加入到TRIE中 
    for(int i=head[x];i;i=edge[i].pre)
	{
        if(node==fa)continue;
        dis[node]=dis[x]^edge[i].f;
        //求出权值，等下加入到TRIE中 
        dfs(node,x);
    }
}
inline int find(int x)
{
    int ans=0,p=0;
    for(int i=31;i>=0;i--){
        bool f=((x&(1<<i))==0);
        if(trie[p][f])
		{
            ans+=1<<i;
            p=trie[p][f];
        }
        else p=trie[p][f^1];
    }
    return ans;
}
int ans;
int main(){
    scanf("%d",&n);
    for(int i=1;i<n;i++)
	{
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);
        add(y,x,z);
    }
    dfs(1,0);//遍历树 
    for(int i=1;i<=n;i++)
	{
        ans=max(ans,find(dis[i]));//查询，并求最大值 
    }
    printf("%d
",ans);
    return 0;
}