Text return string of small MB

Description

Small M prefers to return to the text string, which is a string that reads the same from left to right and from right to left.

Now the small x finds the small M and gives it a string S, hoping that the small M will convert it into a return string.

Small m can perform the following three types of operations:

1. Add add a character anywhere in the string
2. Erase deletes a character
3. Change one letter to another to change a character to another

However, each operation can only be valid for specified characters. For example, small m can only allow erase 'A', add 'B', change 'C' to 'D', and no other operations can be used. There is no free lunch in the world, and every small M operation has a certain cost.

Input Format

The first line is a string that represents the string to be modified by the small M ( Length ≤ 50 )

The number N in the second row indicates the operations that small m can use ( N ≤ 50 )

The next n rows have one operation per row, which can be used in the following three forms:

"Add C x": the cost of adding a character C to a string is X.

"Erase C x": indicates that it takes X to delete a character C from the string.

"Change C1 C2 X": the cost of converting a character C1 to C2 is X.

Note: "Change C1 C2 X" cannot change C2 to C1

Yes X ≤ 100000, C1 <> C2 ,
And any two rows have different operations.

Output Format

A row with one number indicates that the minimum cost of converting the initial string to the input string is M. If not, output-1

Sample Input

caaaaaab6change b a 100000change c a 100000change c d 50000change b e 50000erase d 50000erase e 49999

Sample output

199999

Problem: several groups of data cannot be used. I don't know why ···. Paste the code first and then think about it.

# Include <cstdio> # include <cstring> # include <algorithm> using namespace STD; char s [100]; int ch [40] [40], er [40], add [40]; int DP [100] [100]; const int INF = 999999999; void Init () {for (INT I = 0; I <26; I ++) {er [I] = add [I] = inf; For (Int J = 0; j <26; j ++) ch [I] [J] = (I = J? 0: INF) ;}} void Floyd () {int I, j, k; For (k = 0; k <26; k ++) for (I = 0; I <26; I ++) for (j = 0; j <26; j ++) CH [I] [J] = min (CH [I] [J], ch [I] [k] + CH [k] [J]); for (I = 0; I <26; I ++) for (j = 0; j <26; j ++) if (CH [I] [J]! = Inf) {er [I] = min (ER [I], CH [I] [J] + er [J]); add [I] = min (add [I], CH [I] [J] + Add [J]) ;}} int operdp () {int I, J; int Len = strlen (s); memset (DP, 0, sizeof (DP); for (I = 0; I <Len; I ++) for (j = I + 1; j <Len; j ++) DP [I] [J] = inf; for (I = len-2; I> = 0; I --) {for (j = I + 1; j <Len; j ++) {If (s [I]! = S [J]) {If (ER [s [I]-'a']! = Inf) // Delete s [I] DP [I] [J] = min (DP [I + 1] [J] + er [s [I]-'A' on the left end' ], DP [I] [J]); If (add [s [I]-'a']! = Inf) // Add s [I] DP [I] [J] = min on the right end (DP [I + 1] [J] + Add [s [I]-'a'], DP [I] [J]); If (ER [s [J]-'a']! = Inf) // Delete s [J] DP [I] [J] = min (DP [I] [J-1] + er [s [J]-'a'] on the right end, DP [I] [J]); If (add [s [J]-'a']! = Inf) // Add s [J] DP [I] [J] = min (DP [I] [J-1] + Add [s [J]-'a'] on the left end, DP [I] [J]); If (CH [s [I]-'a'] [s [J]-'a']! = Inf) DP [I] [J] = min (DP [I + 1] [J-1] + CH [s [I]-'a'] [s [J]-'A' ], DP [I] [J]); // change the left end to the same as the right end if (CH [s [J]-'a'] [s [I]-'a']! = Inf) DP [I] [J] = min (DP [I + 1] [J-1] + CH [s [J]-'a'] [s [I]-'A' ], DP [I] [J]); // change the right end to the same as the left end} else {If (ER [s [I]-'a']! = Inf) // Delete s [I] DP [I] [J] = min (DP [I + 1] [J] + er [s [I]-'A' on the left end' ], DP [I] [J]); If (ER [s [I]-'a']! = Inf) // Add s [I] DP [I] [J] = min on the right end (DP [I + 1] [J] + Add [s [I]-'a'], DP [I] [J]); If (ER [s [J]-'a']! = Inf) // Delete s [J] DP [I] [J] = min (DP [I] [J-1] + er [s [J]-'a'] on the right end, DP [I] [J]); If (ER [s [J]-'a']! = Inf) // Add s [J] DP [I] [J] = min (DP [I] [J-1] + Add [s [J]-'a'] on the left end, DP [I] [J]); DP [I] [J] = min (DP [I + 1] [J-1], DP [I] [J]) ;}}if (DP [0] [len-1] = inf) Return-1; return DP [0] [len-1];} int main () {int N, V; char c1 [3], C2 [3], clerk [100]; while (scanf ("% s", S )! = EOF) {Init (); scanf ("% d", & N); While (n --) {scanf ("% s", seconds ); if (strcmp (separator, "change") = 0) {scanf ("% S % d", C1, C2, & V ); ch [c1 [0]-'a'] [c2 [0]-'a'] = V;} else if (strcmp (separator, "erase") = 0) {scanf ("% S % d", C1, & V); er [c1 [0]-'a'] = V ;} else {scanf ("% S % d", C1, & V); add [c1 [0]-'a'] = V ;}} Floyd (); printf ("% d \ n", operdp ();} return 0 ;}