Codeforces 706 C. Hard problem (dp)

题目链接：http://codeforces.com/problemset/problem/706/C

给你n个字符串，可以反转任意一个字符串，反转每个字符串都有其对应的花费ci。

经过操作后是否能满足字符串str[i]>=str[i-1]，能就输出最小花费，不能输出-1。

dp[i][0] 表示不反转i的最小花费(str[i] >= str[i - 1] || str[i] >= reverse(str[i - 1]))

dp[i][1] 则表示反转i的最小花费...

初始dp[1][0] = 0, dp[1][1] = c[1]

要是dp[i][0/1]等于-1 就不能转移了

代码写的有点糟糕，还是太渣...

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 1e5 + ;
 string str[N];
 LL num[N], inf = 1e15;
 LL dp[N][];
 
 int main()
 {
     ios::sync_with_stdio(false);
     int n;
     cin >> n;
     for(int i = ; i <= n; ++i)
         cin >> num[i];
     for(int i = ; i <= n; ++i)
         cin >> str[i];
     int ok = ;
     memset(dp, -, sizeof(dp));
     dp[][] = , dp[][] = num[];
     for(int i = ; i <= n; ++i) {
         string str1 = str[i - ]; //未反转
         reverse(str[i - ].begin(), str[i - ].end());
         string str2 = str[i]; //未反转
         reverse(str[i].begin(), str[i].end());
         if(dp[i - ][] == - && dp[i - ][] == -) {
             ok = -; //不行了
             break;
         }
         if(dp[i - ][] != -) {
             if(str2 >= str1) {
                 dp[i][] = dp[i - ][];
             }
             if(str[i] >= str1) {
                 dp[i][] = dp[i - ][] + num[i];
             }
         }
         if(dp[i - ][] != -) {
             if(str2 >= str[i - ]) {
                 dp[i][] = min(dp[i - ][], dp[i][] == - ? inf : dp[i][]);
             }
             if(str[i] >= str[i - ]) {
                 dp[i][] = min(dp[i - ][] + num[i], dp[i][] == - ? inf : dp[i][]);
             }
         }
         str[i] = str2; //赋值未反转
     }
     if(ok == -) {
         cout << - << endl;
     }
     else if(dp[n][] != - && dp[n][] != -) {
         cout << min(dp[n][], dp[n][]) << endl;
     }
     else if(dp[n][] != -) {
         cout << dp[n][] << endl;
     }
     else if(dp[n][] != -) {
         cout << dp[n][] << endl;
     }
     else {
         cout << - << endl;
     }
     return ;
 }