https://codeforces.com/contest/1499/problem/C
思路:
横纵分开.而且是对称的.所以说,不管你的序列咋样,奇数肯定是一串,偶数是一串.
比如 1 2 3 4 5,不管你咋转,1 3 5是一个集合,24是一个集合。
于是On枚举,维护一个最小值和累加的前缀,最小的走尽可能多的路,剩下的前面的就都走一步。On枚举出最小值
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e5+100;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL a[maxn];
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL t;cin>>t;
while(t--){
LL n;cin>>n;
for(LL i=1;i<=n;i++) cin>>a[i];
LL ans1=1e18;
///先上再右
ans1=n*a[1]+n*a[2];
/// debug(ans1);
LL jmin=a[1];LL omin=a[2];
LL sum1=a[1];LL sum2=a[2];
LL num1=1;LL num2=1;
for(LL i=3;i<=n;i++){
if(i&1){
jmin=min(jmin,a[i]);
sum1+=a[i];
num1++;
ans1=min(ans1,sum1-jmin+ (n-(num1-1) )*jmin+sum2-omin+(n-(num2-1) )*omin );
/// cout<<"i="<<i<<" "<<"ans1="<<ans1<<"\n";
}
else if(i%2==0){
omin=min(omin,a[i]);
sum2+=a[i];
num2++;
ans1=min(ans1,sum2-omin+(n-(num2-1))*omin+sum1-jmin+(n-(num1-1))*jmin );
/// cout<<"i="<<i<<" "<<"ans1="<<ans1<<"\n";
}
}
cout<<ans1<<"\n";
///先右再上
LL ans2=1e18;
}
return 0;
}