bzoj1954 The xor-longest path

Description

 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....

  因为这道题求的是树上的最大异或和路径。我们知道对于树上路径，要么是一条链，要么是对于一个节点的两个不同的子链。 我们记录树上每个点到根节点的亦或和。对于每个异或和的二进制建一颗trie树。 然后贪心，从trie根节点向下，尽量使跑到的边和原本的异或和不同，因为这样它们再次亦或才最大。

#include 
#include 
#include 
#include 
#include 
#include 
#define REP(i,k,n) for(int i=k;i<=n;i++)
#define in(a) a=read()
#define MAXN 300010
using namespace std;
inline int read(){
    int x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())
        if(ch=='-')
            f=-1;
    for(;isdigit(ch);ch=getchar())
        x=x*10+ch-'0';
    return x*f;
}
queue <int> Q;
int n,cnt=0,ans;
int vis[MAXN],sum[MAXN];
int tree[6333333][2];
int total=0,head[MAXN],nxt[MAXN],to[MAXN],val[MAXN];
inline void adl(int a,int b,int c){
    total++;
    to[total]=b;
    val[total]=c;
    nxt[total]=head[a];
    head[a]=total;
    return ;
}
inline void bfs(){
    Q.push(1);
    vis[1]=1;
    while(!Q.empty()){
        int u=Q.front();
        Q.pop();
        for(int e=head[u];e;e=nxt[e])
            if(!vis[to[e]]){
                vis[to[e]]=1;
                sum[to[e]]=sum[u]^val[e];
                Q.push(to[e]);
            }
    }
    return ;
}
inline void insert(int x){
    int u=0;
    for(int i=30;i>=0;i--){
        int t=x&(1<<i);
        t=(t>>i);
        if(!tree[u][t])  tree[u][t]=++cnt;
        u=tree[u][t];
    }
    return ;
}
inline int query(int x){
    int u=0,tmp=0;
    for(int i=30;i>=0;i--){
        int t=x&(1<<i);
        t=(t>>i);
        if(tree[u][t^1])  u=tree[u][t^1],tmp+=(1<<i);
        else  u=tree[u][t];
    }
    return tmp;
}
int main(){
    in(n);
    int a,b,c;
    REP(i,1,n-1) in(a),in(b),in(c),adl(a,b,c),adl(b,a,c);
    bfs();
    REP(i,1,n)  insert(sum[i]);
    REP(i,1,n){
      ans=max(ans,query(sum[i]));
    }
    cout<<ans;
    return 0;
}