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=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }
 struct edge{int go,next;ll w;}e[*maxn];
 int n,tot,cnt,head[maxn],t[maxm][];
 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=;
     for3(i,,)
     {
         int j=x>>i&;
         if(!t[now][j])t[now][j]=++cnt;
         now=t[now][j];
     }
 }
 inline ll query(ll x)
 {
     int now=;ll ret=;
     for3(i,,)
     {
         int j=x>>i&^;
         if(t[now][j])ret|=(ll)<<i;else j^=;
         now=t[now][j];
     }
     return ret;
 }

 int main()

 {

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

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

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

     return ;

 }  

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

鸣谢陶文博