poj 3764 字典树

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7332		Accepted: 1555

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

题意：

给出一棵树，求这棵树中的最大的异或路径。

代码：

//预处理出来每个点到根的异或值sxor，然后u,v之间的异或路径值就是sxor[u]^sxor[v],lca就消去了。然后把每个点的sxor值插入字典
//树（二进制字典树），枚举每个点贪心的找他的最大异或路径。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=200000;
int head[MAXN*2+9],tot,sxor[MAXN*2+9],sz,nod[MAXN*2+9][2],ans;
struct Edge
{
    int to,w,next;
}edge[MAXN*4+9];
void init()
{
    memset(sxor,0,sizeof(sxor));
    memset(head,-1,sizeof(head));
    nod[0][0]=nod[0][1]=0;
    tot=0;
    sz=0;ans=0;
}
void add(int x,int y,int z)
{
    edge[tot].to=y;
    edge[tot].w=z;
    edge[tot].next=head[x];
    head[x]=tot++;
    edge[tot].to=x;
    edge[tot].w=z;
    edge[tot].next=head[y];
    head[y]=tot++;
}
void insert(int x)
{
    int rt=0;
    for(int i=30;i>=0;i--){
        int id=(x>>i)&1;
        if(nod[rt][id]==0){
            nod[rt][id]=++sz;
            nod[sz][0]=nod[sz][1]=0;
        }
        rt=nod[rt][id];
    }
}
void dfs(int x,int fa,int sum)
{
    for(int i=head[x];i!=-1;i=edge[i].next){
        int y=edge[i].to;
        if(y==fa) continue;
        sxor[y]=(sum^edge[i].w);
        dfs(y,x,sum^edge[i].w);
    }
}
void solve(int x)
{
    int sum=0,rt=0;
    for(int i=30;i>=0;i--){
        int id=(x>>i)&1;
        if(nod[rt][!id]){
            sum|=(1<<i);
            rt=nod[rt][!id];
        }else if(nod[rt][id]) rt=nod[rt][id];
        else break;
    }
    ans=max(ans,sum);
}
int main()
{
    int n;
    while(scanf("%d",&n)==1){
        int x,y,z;
        init();
        for(int i=1;i<n;i++){
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);
        }
        dfs(0,-1,0);
        for(int i=0;i<n;i++){
            solve(sxor[i]);
            insert(sxor[i]);
        }
        printf("%d
",ans);
    }
    return 0;
}