Codeforces Round #367 (Div. 2)C

C. Hard problem
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasiliy is fond of solving different tasks. Today he found one he wasn’t able to solve himself, so he asks you to help.

Vasiliy is given n strings consisting of lowercase English letters. He wants them to be sorted in lexicographical order (as in the dictionary), but he is not allowed to swap any of them. The only operation he is allowed to do is to reverse any of them (first character becomes last, second becomes one before last and so on).

To reverse the i-th string Vasiliy has to spent ci units of energy. He is interested in the minimum amount of energy he has to spent in order to have strings sorted in lexicographical order.

String A is lexicographically smaller than string B if it is shorter than B (|A| < |B|) and is its prefix, or if none of them is a prefix of the other and at the first position where they differ character in A is smaller than the character in B.

For the purpose of this problem, two equal strings nearby do not break the condition of sequence being sorted lexicographically.

Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of strings.

The second line contains n integers ci (0 ≤ ci ≤ 109), the i-th of them is equal to the amount of energy Vasiliy has to spent in order to reverse the i-th string.

Then follow n lines, each containing a string consisting of lowercase English letters. The total length of these strings doesn’t exceed 100 000.

Output
If it is impossible to reverse some of the strings such that they will be located in lexicographical order, print  - 1. Otherwise, print the minimum total amount of energy Vasiliy has to spent.

Examples
input
2
1 2
ba
ac
output
1
input
3
1 3 1
aa
ba
ac
output
1
input
2
5 5
bbb
aaa
output
-1
input
2
3 3
aaa
aa
output
-1

思路：
初始化dp为无穷大，然后只有当满足字典序才能去比较min，0和1分别为不转置和转置后的字串。

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int MAXN=100005;LL dp[MAXN][2];LL a[MAXN];string s1[MAXN],s2[MAXN];int n;void init(){    for(int i=0;i<=n;i++)        dp[i][0]=dp[i][1]=1e18;}int main(){    cin>>n;    init();    for(int i=1;i<=n;i++)scanf("%I64d",a+i);    for(int i=1;i<=n;i++){        cin>>s1[i];        s2[i]=s1[i];        reverse(s2[i].begin(),s2[i].end());    }    dp[1][0]=0;    dp[1][1]=a[1];    for(int i=2;i<=n;i++){        if(s1[i]>=s1[i-1])dp[i][0]=min(dp[i][0],dp[i-1][0]);        if(s1[i]>=s2[i-1])dp[i][0]=min(dp[i][0],dp[i-1][1]);        if(s2[i]>=s1[i-1])dp[i][1]=min(dp[i][1],dp[i-1][0]+a[i]);        if(s2[i]>=s2[i-1])dp[i][1]=min(dp[i][1],dp[i-1][1]+a[i]);    }    LL ans=min(dp[n][0],dp[n][1]);    if(ans==1e18)cout<<-1<<endl;    else cout<<ans<<endl;    return 0;}