2019hdu多校 AND Minimum Spanning Tree

题目链接:Click here

题目大意:两个点之间的边权为编号按位与的值，求最小生成树，方案要字典序最小

Solution:

一道不难的构造题，每个点连向他取反后的lowbit值，这样边权为0，若lowbit值大于n，则连1

这样自构造出来的必然是最小生成树，且满足字典序最小

Code:

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+1;
int n,ans=0,a[N];
int lowbit(int x){return x&(-x);}
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
    return x*f;
}
void solve(){
    int n=read(),ans=0;
    for(int i=2;i<=n;i++){
        int x=lowbit(~i);
        a[i]=x<=n?x:1;
        ans+=(a[i]&i);
    }printf("%d
",ans);
    for(int i=2;i<n;i++) printf("%d ",a[i]);
    printf("%d
",a[n]);
}
int main(){
    int t=read();
    while(t--) solve();
    return 0;
}