P4551 (poj3764) Maximum exclusive or path

Description

Given a weighted tree with N points, The subscripts start from 1 to n. Find two nodes in the tree and find the longest exclusive or path.

An exclusive or path refers to the exclusive or of all edge weights in a unique path between two nodes.

Individuals:

First, I strongly recommend 01 dictionary tree (trie). This is a solution.XOR Problems.

When we look for the maximum variance or value, we generally look down from the highest bit to the low bit.

   eg:   1000(2)=8(10)         0111(2)=7(10)

Obviously, as long as my highest bit is 1, unless you and I have the same highest bit, I am bigger than you.

According to the sum of the mathematical proportional series, we can see that
8 = 2 ^ 3, 7 = 2 ^ 3-1

So we canGreedy to find the current node ^ 1

01 The dictionary tree method is similar to the trie tree method. For this question,

Process: 1. Create a graph and run DFS to find the XOR value from each node to the root node. 2. Build 01trie again to implement our greed.

I have talked a lot about this on the Internet. I want to learn Baidu.

------- Code ---------

#include<bits/stdc++.h>#define IL inline#define RI register int#define maxn 100008int trie[maxn*31][2],xo[maxn],ans,rt;int val[maxn],n,head[maxn],tot;struct code{int u,v,w;}edge[maxn<<1];IL void read(int &x){    int f=1;x=0;char s=getchar();    while(s>'9'||s<'0'){if(s=='-')f=-1;s=getchar();}    while(s<='9'&&s>='0'){x=x*10+s-'0';s=getchar();}    x*=f;}IL void add(int x,int y,int z){    edge[++tot].u=head[x];    edge[tot].v=y;    edge[tot].w=z;    head[x]=tot;    edge[++tot].u=head[y];    edge[tot].v=x;    edge[tot].w=z;    head[y]=tot;}IL void build_trie(int x,int rt){    for(RI i=1<<30;i;i>>=1)    {        bool c=x&i;        if(!trie[rt][c])trie[rt][c]=++tot;        rt=trie[rt][c];    }}IL int query(int x,int rt){    int ans=0;    for(RI i=1<<30;i;i>>=1)    {        bool c=x&i;        if(trie[rt][c^1])ans+=i,rt=trie[rt][c^1];        else rt=trie[rt][c];    }    return ans;}IL void dfs(int u,int fa){    for(RI i=head[u];i;i=edge[i].u)    {        if(edge[i].v!=fa)        {            xo[edge[i].v]=xo[u]^edge[i].w;            dfs(edge[i].v,u);        }    }}int main(){    read(n);    for(RI i=1,u,v,w;i<n;i++)read(u),read(v),read(w),add(u,v,w);    dfs(1,0);    for(RI i=1;i<=n;i++)build_trie(xo[i],rt);    for(RI i=1;i<=n;i++)ans=std::max(ans,query(xo[i],rt));    printf("%d",ans);}

P4551 (poj3764) Maximum exclusive or path