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; }
时间: 2024-10-18 20:14:28