#include <iostream>
#include <sstream>
#include <iomanip>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <numeric>
#include <iterator>
#include <functional>
#include <complex>
#include <queue>
#include <stack>
#include <cmath>
#include <cassert>
#include <cstring>
using namespace std;
typedef long long LL;
typedef complex<double> CMP;
class QuickT9 { public:
int minimumPressings(vector <string> t9, string word)
{
map< string, set<string> > D2U;
for(int i=0; i<t9.size(); ++i) {
stringstream sin(t9[i]);
for(string u;sin>>u;) {
string d;
for(int i=0; i<u.size(); ++i) {
if(u[i]<='c') d+='2';
else if(u[i]<='f') d+='3';
else if(u[i]<='i') d+='4';
else if(u[i]<='l') d+='5';
else if(u[i]<='o') d+='6';
else if(u[i]<='s') d+='7';
else if(u[i]<='v') d+='8';
else if(u[i]<='z') d+='9';
D2U[d].insert( u.substr(0,i+1) );
}
}
}
return rec(D2U, word, 0, "", "");
}
// Now, the state is F=word[0..i), D=D, U=U. What is the minimal number of steps from here?
// It is computed by simply implementing the problem statement, with memoization.
map<pair<int,string>, int> memo;
int rec( map< string, set<string> >& D2U, const string& word, int i, const string& D, const string& U )
{
if( i == word.size() ) return 0;
if( i < 0 ) return -1;
pair<int,string> key(i,U);
if( memo.count(key) )
return memo[key];
memo[key] = -1; // avoid cyclic dependency
int best = -1;
#define UPDATE(s,k) { if(s!=-1 && (best==-1 || s+(k)<best)) best=s+(k); }
// try RIGHT
{
int cp = 0;
while( i+cp<word.size() && cp<U.size() && word[i+cp] == U[cp] )
++cp;
int k = U.size()-cp;
int s = rec(D2U, word, i+U.size()-k, "", "");
UPDATE(s,k+1);
}
// try C
{
int cp = 0;
while( i+cp<word.size() && cp<U.size() && word[i+cp] == U[cp] )
++cp;
int k = U.size()-cp;
if(k==0) k=1; // the only difference is here
int s = rec(D2U, word, i+U.size()-k, "", "");
UPDATE(s,k);
}
// try *
if( !D.empty() ) {
set<string>& Us = D2U[D];
set<string>::iterator it = Us.find(U);
++it;
if( it == Us.end() )
it = Us.begin();
int s = rec(D2U, word, i, D, *it);
UPDATE(s,1);
}
// try 2-9
for(char c='2'; c<='9'; ++c)
{
string nextD = D + c;
if( D2U.count(nextD) ) {
set<string>& Us = D2U[nextD];
int s = rec(D2U, word, i, nextD, *Us.begin());
UPDATE(s,1);
}
}
// done
return memo[key] = best;
}
};
// BEGIN CUT HERE
#include <ctime>
double start_time; string timer()
{ ostringstream os; os << " (" << int((clock()-start_time)/CLOCKS_PER_SEC*1000) << " msec)"; return os.str(); }
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v)
{ os << "{ ";
for(typename vector<T>::const_iterator it=v.begin(); it!=v.end(); ++it)
os << '\"' << *it << '\"' << (it+1==v.end() ? "" : ", "); os << " }"; return os; }
void verify_case(const int& Expected, const int& Received) {
bool ok = (Expected == Received);
if(ok) cerr << "PASSED" << timer() << endl; else { cerr << "FAILED" << timer() << endl;
cerr << "\to: \"" << Expected << '\"' << endl << "\tx: \"" << Received << '\"' << endl; } }
#define CASE(N) {cerr << "Test Case #" << N << "..." << flush; start_time=clock();
#define END verify_case(_, QuickT9().minimumPressings(t9, word));}
int main(){
CASE(0)
string t9_[] = {"aae", "bab", "abad", "bdbd", "beta"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "babe";
int _ = 9;
END
CASE(1)
string t9_[] = {"ann","ie"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "annie";
int _ = 7;
END
CASE(2)
string t9_[] = {"ann","amm"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "annie";
int _ = -1;
END
CASE(3)
string t9_[] = {"aaa aab","aac aba abb ccca"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "aba";
int _ = 6;
END
CASE(4)
string t9_[] = {"acac aba aaab","aab aa baa","bba bacade abb","baba"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "abbaca";
int _ = 10;
END
CASE(5)
string t9_[] = {"aaa aab aac","aba abb","ccca"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "ccc";
int _ = 5;
END
CASE(6)
string t9_[] = {"abc"};
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = "a";
int _ = 2;
END
/*
CASE(7)
string t9_[] = ;
vector <string> t9(t9_, t9_+sizeof(t9_)/sizeof(*t9_));
string word = ;
int _ = ;
END
*/
}
// END CUT HERE