CF 706C Hard problem

DP

dp[i][j]，i、j分别表示到第i组为止，第i组为j状态的最小花费，其中j有两种状态：0表示不翻转，1表示翻转。

#include <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

#define MAXN 100000
#define INF 0x3f3f3f3f3f3f3f3f

int n;
long long c[MAXN+10], dp[MAXN+10][2];
vector<string> str;

int main() {
    str.clear();
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> c[i];
    str.push_back("");
    str.push_back("");
    for (int i = 1; i <= n; ++i) {
        string s;
        s.clear();
        cin >> s;
        str.push_back(s);
        reverse(s.begin(), s.end());
        str.push_back(s);
    }
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j < 2; ++j)
            dp[i][j] = INF;
    dp[0][0] = dp[0][1] = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < 2; ++j) {
            for (int k = 0; k < 2; ++k) {
                if (dp[i-1][j] != INF && str[2*(i-1)+j] <= str[2*i+k]) {
                    dp[i][k] = min(dp[i][k], dp[i-1][j]+k*c[i]);
                }
            }
        }
    }
    long long res = min(dp[n][0],dp[n][1]);
    if (res == INF) cout << -1 << endl;
    else cout << res << endl;
    return 0;
}