BZOJ1954: Pku3764 The xor-longest Path

题解：

在树上i到j的异或和可以直接转化为i到根的异或和^j到根的异或和。

所以我们把每个点到根的异或和处理出来放到trie里面，再把每个点放进去跑一遍即可。

代码：

#include<cstdio>

#include<cstdlib>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<string>

#define inf 1000000000

#define maxn 100000+5

#define maxm 4000000+5

#define eps 1e-10

#define ll long long

#define pa pair<int,int>

#define for0(i,n) for(int i=0;i<=(n);i++)

#define for1(i,n) for(int i=1;i<=(n);i++)

#define for2(i,x,y) for(int i=(x);i<=(y);i++)

#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define for4(i,x) for(int i=head[x],y;i;i=e[i].next)

#define mod 1000000007

using namespace std;

inline 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=10*x+ch-'0';ch=getchar();}

    return x*f;

}
struct edge{int go,next;ll w;}e[2*maxn];
int n,tot,cnt,head[maxn],t[maxm][2];
ll a[maxn];
inline void insert(int x,int y,ll z)
{
    e[++tot]=(edge){y,head[x],z};head[x]=tot;
    e[++tot]=(edge){x,head[y],z};head[y]=tot;
}
inline void dfs(int x,int fa)
{
    for4(i,x)if((y=e[i].go)!=fa)
    {
        a[y]=a[x]^e[i].w;
        dfs(y,x);
    }
}
inline void add(ll x)
{
    int now=1;
    for3(i,62,0)
    {
        int j=x>>i&1;
        if(!t[now][j])t[now][j]=++cnt;
        now=t[now][j];
    }
}
inline ll query(ll x)
{
    int now=1;ll ret=0;
    for3(i,62,0)
    {
        int j=x>>i&1^1;
        if(t[now][j])ret|=(ll)1<<i;else j^=1;
        now=t[now][j];
    }
    return ret;
}

int main()

{

    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

    n=read();
    for1(i,n-1){int x=read(),y=read(),z=read();insert(x,y,z);}
    dfs(1,0);
    cnt=1;
    for1(i,n)add(a[i]);
    ll ans=0;
    for1(i,n)ans=max(ans,query(a[i]));
    cout<<ans<<endl;

    return 0;

}  

1954: Pku3764 The xor-longest Path

Time Limit: 1 Sec  Memory Limit: 64 MB
Submit: 383  Solved: 161
[Submit][Status]

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....

Source

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