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Solution

简单题？
考虑一条路径长什么样子，一定是经过某一个字母环的左上角，那么答案就很简单了。
我们记一个前缀最小值，这样子让他一路走下去一定是最优！
然后扫一遍就好了。

代码实现

/*  mail: mleautomaton@foxmail.com  author: MLEAutoMaton  This Code is made by MLEAutoMaton*/#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<algorithm>#include<queue>#include<set>#include<map>#include<iostream>using namespace std;#define ll long long#define re register#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)inline int gi(){    int f=1,sum=0;char ch=getchar();    while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}    return f*sum;}inline ll gl(){    ll f=1,sum=0;char ch=getchar();    while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}    return f*sum;}ll ans=1e18+10,Min=1e18+10;ll a[400010],sum,n;int main(){    n=gl()+1;    for(int i=1;i<=n;i++)a[i]=gl();reverse(a+1,a+n+1);    for(int i=1;i<=n;i++)    {        Min=min(Min,a[i]);        ans=min(ans,a[i]*(4*(n-i)+1)+sum*2);        sum+=a[i]+Min;    }    printf("%lld\n",ans);    return 0;}