P1032 String Conversion and p1032 String Conversion

P1032 String Conversion and p1032 String Conversion
Description

It is known that there are two strings A, B and A set of string conversion rules (up to 6 Rules ):

A1-> B1

A2-> B2

The meaning of the rule is: the sub-string A1 in A $ can be changed to B1, A2 can be changed to B2 ....

For example, A = 'abc' B = 'xyz'

The conversion rules are as follows:

'Abc'-> 'xu ''ud'-> 'y' y '-> 'yz'

At this time, A can undergo A series of transformations to B, and the transformation process is:

'Abcd'-> 'xud '-> 'xy'-> 'xyz'

Three transformations are performed to convert A to B.

Input/Output Format Input Format:

Enter the keyboard file name. The file format is as follows:

A BA1 B1 \

A2 B2 |-> conversion rules

....../

The maximum length of all strings is 20.

Output Format:

Output to the screen. The format is as follows:

If A can be converted to B within 10 steps (including 10 steps), the minimum number of transformation steps is output; otherwise, the output is "NO ANSWER! "

Input and Output sample Input example #1:

abcd xyzabc xuud yy yz

Output sample #1:

3

I have done this before,
However, it is the last vertex that is used only when a table is created.
I redo it today.
To realize the truth:
Never over-reliance on STL!

 1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 #include<cstring> 5 #include<cstdlib> 6 #include<map> 7 using namespace std; 8 string bg,ed; 9 struct guize10 {11     string a,b;12 }gz[1001];13 struct node14 {15     string zfc;16     int step;17 }now,nxt;18 int num=1;19 map<string,bool>mp;20 int tot=0;21 void bfs()22 {23     queue<node>q;24     now.zfc=bg;now.step=0;25     q.push(now);26     while(q.size()!=0)27     {28         node p=q.front();29         //cout<<p.zfc<<"*4654654"<<endl;30         if(p.zfc==ed)31         {32             printf("%d",p.step);33             exit(0);34         }35         if(p.step>10)36         {37             printf("NO ANSWER!");38             exit(0);39         }40         q.pop();41         mp[p.zfc]=1;42         for(int i=1;i<=num-1;i++)43         {44             string tmp=p.zfc;45             //int where=p.zfc.find(gz[i].a);46             for(int j=0;j<p.zfc.length();j++)47             {48                 if(p.zfc[j]==gz[i].a[0])49                 {50                     int flag=0;51                     for(int k=0;k<gz[i].a.length();k++)52                     {53                         tot++;54                         if(tot>=4384380)55                         {56                             printf("NO ANSWER!");57                             exit(0);58                         }59                         if(p.zfc[j+k]!=gz[i].a[k])60                         {61                             flag=1;break;62                         }63                         64                         65                     }66                     if(flag==0)67                     {68                         int where=j;69                         nxt.zfc=p.zfc.replace(where,gz[i].a.length(),gz[i].b);70                         p.zfc=tmp;71                         if(mp[nxt.zfc]==0)72                         {73                             74                             mp[nxt.zfc]=1;75                             nxt.step=p.step+1;76                             //cout<<p.zfc<<"****"<<nxt.zfc<<"***"<<nxt.step<<endl;77                             q.push(nxt);78                         }79                     }80                 }81             }82         }83     }84 }85 int main()86 {87     freopen("string.in","r",stdin);88     freopen("string.out","w",stdout);89     cin>>bg>>ed;90     while(cin>>gz[num].a>>gz[num].b)91     num++;92     //for(int i=1;i<=num-1;i++)93     //    cout<<gz[i].a<<"+++"<<gz[i].b<<endl;94     bfs();95     printf("NO ANSWER!");96     return 0;97 }