C. Minimum Grid Path(思维+枚举+贪心)

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<<"
";
        }
        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<<"
";
        }

    }
    cout<<ans1<<"
";
    ///先右再上
    LL ans2=1e18;

  }
return 0;
}