CodeForces 706C Hard problem (水DP)

题意：对于给定的n个字符串，可以花费a[i]  将其倒序，问是否可以将其排成从大到小的字典序，且花费最小是多少。

析：很明显的水DP，如果不是水DP，我也不会做。。。。

这个就要二维，d[2][maxn]，d[0][i]表示第 i 个不反转是最小花费，d[1][i]表示第 i 个反转最小花费，那么剩下的就很简单了么，

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <stack>
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 100000000000000000;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
int n, m;
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
inline LL Max(LL a, LL b){  return a < b ? b : a; }
inline LL Min(LL a, LL b){  return a > b ? b : a; }
int a[maxn];
vector<string> v1;
vector<string> v2;
LL d[2][maxn];

int main(){
    while(scanf("%d", &n) == 1){
        for(int i = 0; i < n; ++i)  scanf("%d", &a[i]);
        string s;
        v1.clear();  v2.clear();
        for(int i = 0; i < n; ++i){
            cin >> s;
            v1.push_back(s);
            reverse(s.begin(), s.end());
            v2.push_back(s);
        }
        fill(d[0], d[0]+n, LNF);
        fill(d[1], d[1]+n, LNF);
        d[0][0] = 0, d[1][0] = a[0];
        for(int i = 1; i < n; ++i){
            if(v1[i-1] <= v1[i]) d[0][i] = Min(d[0][i], d[0][i-1]);
            if(v1[i-1] <= v2[i]) d[1][i] = Min(d[1][i], d[0][i-1]+a[i]);
            if(v2[i-1] <= v1[i]) d[0][i] = Min(d[0][i], d[1][i-1]);
            if(v2[i-1] <= v2[i]) d[1][i] = Min(d[1][i], d[1][i-1]+a[i]);
            if(d[1][i] == LNF && d[0][i] == LNF)   break;
        }
        LL ans = Min(d[0][n-1], d[1][n-1]);
        printf("%I64d\n", ans == LNF ? -1 : ans);
    }
    return 0;
}