最短路

做法：

把二进制中的每一位看作是一个点，改位为1就建立长度为的边
跑dij即可

代码

// Problem: 最短路
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11233/D
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<ll,int> PII;
typedef unsigned long long ull;
typedef long double ld;
const int N=100040;
const ll INF=1e18;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

struct edge{
    int to; //连接的节点
    ll cost; //边长
};

vector<edge> g[N];
ll dis[N];
bool vis[N];
int n;

void init(){
    for(int i=1;i<=n+30;i++) dis[i]=INF;
    dis[0]=0;
}

void Dijkstra(){
    init();
    priority_queue<PII,vector<PII>,greater<PII> > que; //按first从小到大
    que.push({0,0});
    while(!que.empty()){
        auto t=que.top();que.pop();
        int v=t.second;
        // if(dis[v]<t.first) continue;
        if(vis[v]) continue;
        vis[v]=1;
        for(int i=0;i<g[v].size();i++){
            auto e=g[v][i];
            if(dis[e.to] > dis[v]+e.cost){
                dis[e.to] = dis[v]+e.cost;
                que.push({dis[e.to],e.to});
            }
        }
    }
}

void solve(){
    cin>>n;
    rep(i,0,n-1){
        ll x;cin>>x;
        for(int j=0;j<=30;j++){
            if((x>>j)&1){
                g[i].pb({n+j,x});
                g[n+j].pb({i,x});
            }
        }
    }
    Dijkstra();
    rep(i,0,n-1){
        if(dis[i]!=INF) cout<<dis[i]<<" \n"[i==n-1];
        else cout<<-1<<" \n"[i==n-1];
    }
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}