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...
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <string> 8 #include <vector> 9 #include <cmath> 10 #include <ctime> 11 #include <list> 12 #include <set> 13 #include <map> 14 using namespace std; 15 typedef long long LL; 16 typedef pair <int, int> P; 17 const int N = 1e5 + 5; 18 string str[N]; 19 LL num[N], inf = 1e15; 20 LL dp[N][2]; 21 22 int main() 23 { 24 ios::sync_with_stdio(false); 25 int n; 26 cin >> n; 27 for(int i = 1; i <= n; ++i) 28 cin >> num[i]; 29 for(int i = 1; i <= n; ++i) 30 cin >> str[i]; 31 int ok = 0; 32 memset(dp, -1, sizeof(dp)); 33 dp[1][0] = 0, dp[1][1] = num[1]; 34 for(int i = 2; i <= n; ++i) { 35 string str1 = str[i - 1]; //未反转 36 reverse(str[i - 1].begin(), str[i - 1].end()); 37 string str2 = str[i]; //未反转 38 reverse(str[i].begin(), str[i].end()); 39 if(dp[i - 1][0] == -1 && dp[i - 1][1] == -1) { 40 ok = -1; //不行了 41 break; 42 } 43 if(dp[i - 1][0] != -1) { 44 if(str2 >= str1) { 45 dp[i][0] = dp[i - 1][0]; 46 } 47 if(str[i] >= str1) { 48 dp[i][1] = dp[i - 1][0] + num[i]; 49 } 50 } 51 if(dp[i - 1][1] != -1) { 52 if(str2 >= str[i - 1]) { 53 dp[i][0] = min(dp[i - 1][1], dp[i][0] == -1 ? inf : dp[i][0]); 54 } 55 if(str[i] >= str[i - 1]) { 56 dp[i][1] = min(dp[i - 1][1] + num[i], dp[i][1] == -1 ? inf : dp[i][1]); 57 } 58 } 59 str[i] = str2; //赋值未反转 60 } 61 if(ok == -1) { 62 cout << -1 << endl; 63 } 64 else if(dp[n][0] != -1 && dp[n][1] != -1) { 65 cout << min(dp[n][0], dp[n][1]) << endl; 66 } 67 else if(dp[n][0] != -1) { 68 cout << dp[n][0] << endl; 69 } 70 else if(dp[n][1] != -1) { 71 cout << dp[n][1] << endl; 72 } 73 else { 74 cout << -1 << endl; 75 } 76 return 0; 77 }
