CodeForces - 706C Hard problem（dp+字符串）

题意：有n个字符串，只能将其逆转，不能交换位置，且已知逆转某字符串需要消耗的能量，问将这n个字符串按字典序从小到大排序所需消耗的最少能量。

分析：每个字符串要么逆转，要么不逆转，相邻两个字符串进行比较，从而可得4个状态转移方程。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL c[MAXN];
string s[MAXN][2];
LL dp[MAXN][2];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; ++i){
        scanf("%I64d", &c[i]);
    }
    string tmp;
    for(int i = 0; i < n; ++i){
        cin >> tmp;
        s[i][0] = tmp;
        reverse(tmp.begin(), tmp.end());
        s[i][1] = tmp;
    }
    memset(dp, LL_INF, sizeof dp);
    dp[0][0] = 0, dp[0][1] = c[0];
    bool ok = true;
    for(int i = 1; i < n; ++i){
        if(s[i][0] >= s[i - 1][0]){
            dp[i][0] = min(dp[i][0], dp[i - 1][0]);
        }
        if(s[i][1] >= s[i - 1][0]){
            dp[i][1] = min(dp[i][1], dp[i - 1][0] + c[i]);
        }
        if(s[i][0] >= s[i - 1][1]){
            dp[i][0] = min(dp[i][0], dp[i - 1][1]);
        }
        if(s[i][1] >= s[i - 1][1]){
            dp[i][1] = min(dp[i][1], dp[i - 1][1] + c[i]);
        }
        if(dp[i][0] == LL_INF && dp[i][1] == LL_INF){
            ok = false;
            break;
        }
    }
    if(ok){
        printf("%I64d\n", min(dp[n - 1][0], dp[n - 1][1]));
    }
    else{
        printf("-1\n");
    }
    return 0;
}