POJ3764 The xor-longest Path

 

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。

 

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=200010;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
struct edge{
    int v,nxt,w;
}e[mxn<<1];
int hd[mxn],mct=0;
void add_edge(int u,int v,int w){
    e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return;
}
int t[mxn*10][2],cnt=1;
int n,ans=0;
int f[mxn];
void insert(int x){
    int now=1;
    for(int i=31;i>=0;i--){
        int v=(x>>i)&1;
        if(!t[now][v])t[now][v]=++cnt;
        now=t[now][v];
    }
    return;
}
void query(int x){
    int now=1;
    int res=0;
    for(int i=31;i>=0;i--){
        int v=(x>>i)&1;
        if(t[now][v^1]){
            v^=1;
            res+=(1<<i);
        }
        now=t[now][v];
    }
    ans=max(res,ans);
    return;
}
void DFS(int u,int fa,int num){
    f[u]=num;
    for(int i=hd[u];i;i=e[i].nxt){
        int v=e[i].v;
        if(v==fa)continue;
        DFS(v,u,num^e[i].w);
    }
    return;
}
void init(){
    memset(hd,0,sizeof hd);
    memset(t,0,sizeof t);
//    memset(f,0,sizeof f);
    mct=0;cnt=1;ans=0;
    return;
}
int main(){
    while(scanf("%d",&n)!=EOF){
        init();
        int i,j,u,v,w;
        for(i=1;i<n;i++){
            u=read();v=read();w=read();
            u++;v++;
            add_edge(u,v,w);
            add_edge(v,u,w);
        }
        DFS(1,0,0);
        for(i=1;i<=n;i++){
//            printf("f[%d]:%d\n",i,f[i]);
            query(f[i]);
            insert(f[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}