思路：我们可以开虚拟节点来连接两个结点权值二进制1的个数相同的两个结点，由于1e9内1的个数不超过30，所以开30个虚拟节点，第i个结点连向二进制1的个数为i的点，然后跑dij即可。

代码：

int n, m;
struct d
{
    int u, w;
};

int f(int x){
    int cnt = 0;
    while(x){
        if(x & 1)
            cnt ++;
        x /= 2;
    }
    return cnt;
}
vector<d>e[100100];
int c[100010], dist[100010], st[100010];
const int C = 100010;

void dij(){
    priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>q;
    memset(dist,0x3f,sizeof dist);
    q.push({0, 1});
    dist[1] = 0;
    while(q.size()){
        auto t = q.top();
        q.pop();
        int d = t.first, u = t.second;
        if(st[u])
            continue;
        st[u] = 1;
        for(auto [x,w] : e[u]){
            if(d + w < dist[x]){
                dist[x] = d + w;
                q.push({x,dist[x]});
            }
        }
    }
}

void solve(){
    cin >> n >> m;
    for(int i = 1;i <= n;i ++){
        cin >> c[i];
        e[i].push_back({f(c[i]) + C, c[i]});
        e[f(c[i]) + C].push_back({i, 0});
    }
    for(int i = 0;i < m;i ++){
        int u,v,w;
        cin >> u >> v >> w;
        e[u].push_back({v,w});
        e[v].push_back({u,w});
    }
    dij();
    if(dist[n] == 0x3f3f3f3f)
        cout << "gou yun hao";
    else
        cout << dist[n];
}