Wide Search--transformation class

Wikioi 1099 String transformation

Title Description Description

It is known that there are two strings a$, b$, and a set of rules for the Transformation of strings (up to 6 rules):
a1$-b1$
a2$-b2$
The meaning of the rule is: the substring a1$ in a$ can be transformed to b1$, a2$ can be transformed into b2$ ....
Example: a$= ' ABCD ' b$= ' xyz '
The transformation rules are:
' The ' abc ', ' Xu ', ' ud ', ' y ' y ', ' yz '

At this point, the a$ can go through a series of transformations into b$, the process of which is:
' xyz ', ' xy ', ' xud ', ' ABCD '

A total of three transformations were made, making the a$ transform into b$.

Enter a description Input Description

The input format is as follows:

a$ b$
a1$ b1$ \
a2$ b2$ |-> Transformation Rules
... ... /
The upper limit for all string lengths is 20.

Output description Output Description

If the a$ is converted to b$ in 10 steps (including 10 steps), the minimum number of conversion steps is output, otherwise the output "NO answer!"

Sample input Sample Input

ABCD XYZ
ABC Xu
UD y
Y yz

Sample output Sample Output

3

Data range and Tips Data Size & Hint

Hehe

Idea: Two-way wide search code:

1#include <stdio.h>2#include <string.h>3 structnode4 {5     Chars[ -];6     intDep//Number of Transformations7} list1[5010], list2[5010];8 Chara[7][ -], b[7][ -];9 intN;Ten voidBFS () One { A     intHead1, Tail1, Head2, Tail2, K; -Head1 = Tail1 = head2 = Tail2 =1; -      while(Head1 <= tail1 && head2 <=tail2) the     { -         if(LIST1[HEAD1].DEP + LIST2[HEAD2].DEP >Ten)  -         { -printf"NO answer!\n"); +             return ; -         } +          for(inti =0; I < strlen (LIST1[HEAD1].S); i++) A              for(intj =1; J <= N; J + +) at                 if(strncmp (List1[head1].s + i, A[j], strlen (a[j])) = =0)//find current rules that can be transformed -                 { -tail1++;//move the tail pointer, store the transformed string, the following three for loops for the transformation process -                    for(k =0; K < I; k++)  -LIST1[TAIL1].S[K] =List1[head1].s[k]; -                    for(intL =0; L < strlen (B[j]); l++, k++)  inLIST1[TAIL1].S[K] =B[j][l]; -                    for(intL = i + strlen (a[j]); L <= strlen (LIST1[HEAD1].S); l++, k++) toLIST1[TAIL1].S[K] =List1[head1].s[l]; +LIST1[TAIL1].S[K] =' /';//string plus terminator at end of transform -LIST1[TAIL1].DEP = list1[head1].dep+1; the                    for(k =1; K <= Tail1; k++) *                     if(strcmp (LIST1[TAIL1].S, list2[k].s) = =0)//determine if the current state is connected to a reverse search $                     {Panax Notoginsengprintf"%d\n", LIST1[TAIL1].DEP +LIST2[K].DEP); -                        return ; the                      } +                 } A          for(inti =0; I < strlen (LIST2[HEAD2].S); i++)//Reverse Search Ibid . the              for(intj =1; J <= N; J + +) +                 if(strncmp (List2[head2].s + i, B[j], strlen (b[j])) = =0) -                 { $tail2++; $                    for(k =0; K < I; k++)  -LIST2[TAIL2].S[K] =List2[head2].s[k]; -                    for(intL =0; L < strlen (A[j]); l++, k++)  theLIST2[TAIL2].S[K] =A[j][l]; -                    for(intL = i + strlen (b[j]); L <= strlen (LIST2[HEAD2].S); l++, k++)WuyiLIST2[TAIL2].S[K] =List2[head2].s[l]; theLIST2[TAIL2].S[K] =' /'; -LIST2[TAIL2].DEP = LIST2[HEAD2].DEP +1; Wu                    for(k =1; k <= Tail1; k++) -                     if(strcmp (LIST1[K].S, list2[tail2].s) = =0) About                     { $printf"%d\n", LIST1[K].DEP +LIST2[TAIL2].DEP); -                        return ; -                     } -                 } Ahead1++;  +head2++; the     } -printf"NO answer!\n"); $ } the intMain () the { thescanf"%s%s", list1[1].S, list2[1].s); then =1; -      while(SCANF ("%s%s", a[n],b[n])! =EOF) inn++; then--; thelist1[1].DEP = list2[1].DEP =0; About BFS (); the     return 0; the}

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Wide Search--transformation class