meaning of the title

Given a tree with edge weights, each operation can make all edges of a chain different or the same number as the previous one. The edge weights of each edge are changed to 0 after the least number of operations.
N < = 100000, border weight < = 15

Analysis

We can set the point weight of a point as XOR sum of all the edges connected with it.
Then the point weight of all points is 0 and the edge weight of all edges is 0, which is a necessary and sufficient condition for each operation to change the point weight of two points.
Press dp one time.

Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
#define MIN(x,y) x=min(x,y)
using namespace std;

const int N=100005;
const int inf=1000000000;

int n,a[N],bin[20],cnt[70005],f[70005],w[20];
vector<int> vec[20];

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;
}

void dp(int S)
{
    for (int i=0;i<bin[16];i++) cnt[i]=cnt[i>>1]+(i&1),vec[cnt[i]].push_back(i),f[i]=inf;
    f[S]=0;
    for (int i=16;i>0;i--)
        for (int j=0;j<vec[i].size();j++)
        {
            int s=vec[i][j];
            if (f[s]==inf) continue;
            for (int x=0;x<16;x++)
                for (int y=0;y<16;y++)
                {
                    if (!(s&bin[x])||!(s&bin[y])||x==y) continue;
                    MIN(f[s^bin[x]^bin[x^y]^bin[y]],f[s]+1+((s&bin[x^y])>0));
                }
        }
}

int main()
{
    bin[0]=1;
    for (int i=1;i<=16;i++) bin[i]=bin[i-1]*2;
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read(),z=read();
        a[x+1]^=z;a[y+1]^=z;
    }
    for (int i=1;i<=n;i++) w[a[i]]++;
    int S=0,ans=0;
    for (int i=1;i<=15;i++) ans+=w[i]/2,w[i]%=2,S+=bin[i]*w[i];
    dp(S);
    printf("%d",ans+f[0]);
    return 0;
}