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/*
EGYPT Toolkit for Statistical Machine Translation
Written by Yaser Al-Onaizan, Jan Curin, Michael Jahr, Kevin Knight, John Lafferty, Dan Melamed, David Purdy, Franz Och, Noah Smith, and David Yarowsky.
This program is free software; you can redistribute it and/or
modify it under the terms of the GNU General Public License
as published by the Free Software Foundation; either version 2
of the License, or (at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program; if not, write to the Free Software
Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307,
USA.
*/
/* Noah A. Smith
Dictionary object for dictionary filter in Model 1 training
Dictionary file must be in order (sorted) by Foreign vocab id, but English
vocab ids may be in any order.
9 August 1999
*/
#include "Dictionary.h"
#include <string.h>
Dictionary::Dictionary(const char *filename){
if(!strcmp(filename, "")){
dead = true;
return;
}
dead = false;
cout << "Reading dictionary from: " << filename << '\n';
ifstream dFile(filename);
if(!dFile){
cerr << "ERROR: Can't open dictionary: " << filename << '\n';
exit(1);
}
currindexmin = 0;
currindexmax = 0;
currval = 0;
int p, q;
while((dFile >> p >> q)){
pairs[0].push_back(p);
pairs[1].push_back(q);
}
cout << "Dictionary read; " << pairs[0].size() << " pairs loaded." << '\n';
dFile.close();
}
bool Dictionary::indict(int p, int q){
if(dead) return false;
if(p == 0 && q == 0) return false;
if(currval == p){
for(int i = currindexmin; i <= currindexmax; i++)
if(pairs[1][i] == q) return true;
return false;
}
else{
int begin = 0, end = pairs[0].size() - 1, middle = 0;
unsigned int t;
bool ret = false;
while(begin <= end){
middle = begin + ((end - begin) >> 1);
if(p < pairs[0][middle]) end = middle - 1;
else if(p > pairs[0][middle]) begin = middle + 1;
else{
break;
}
}
t = middle;
while(pairs[0][t] == p )
if(pairs[1][t--] == q) ret = true;
currindexmin = t + 1;
t = middle + 1;
while(pairs[0][t] == p && t < pairs[0].size())
if(pairs[1][t++] == q) ret = true;
currindexmax = t - 1;
currval = p;
return ret;
}
}
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