HDU-1306-String Matching

String Matching

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 776    Accepted Submission(s): 404


[align=left]Problem Description[/align]
It's easy to tell if two words are identical - just check the letters. But how do you tell if two words are almost identical? And how close is "almost"?

There are lots of techniques for approximate word matching. One is to determine the best substring match, which is the number of common letters when the words are compared letter-byletter.

The key to this approach is that the words can overlap in any way. For example, consider the words CAPILLARY and MARSUPIAL. One way to compare them is to overlay them: 

CAPILLARY

MARSUPIAL

There is only one common letter (A). Better is the following overlay:

CAPILLARY

   MARSUPIAL

with two common letters (A and R), but the best is:

  CAPILLARY

MARSUPIAL 

Which has three common letters (P, I and L).

The approximation measure appx(word1, word2) for two words is given by:

common letters * 2

-----------------------------

length(word1) + length(word2)

Thus, for this example, appx(CAPILLARY, MARSUPIAL) = 6 / (9 + 9) = 1/3. Obviously, for any word W appx(W, W) = 1, which is a nice property, while words with no common letters have an appx value of 0.

 

[align=left]Sample Input[/align]

The input for your program will be a series of words, two per line, until the end-of-file flag of -1.

Using the above technique, you are to calculate appx() for the pair of words on the line and print the result. For example:

CAR CART
TURKEY CHICKEN
MONEY POVERTY
ROUGH PESKY
A A
-1

The words will all be uppercase.

 

[align=left]Sample Output[/align]

Print the value for appx() for each pair as a reduced fraction, like this:

appx(CAR,CART) = 6/7
appx(TURKEY,CHICKEN) = 4/13
appx(MONEY,POVERTY) = 1/3
appx(ROUGH,PESKY) = 0
appx(A,A) = 1

 

[align=left]Recommend[/align]
Eddy
 

import java.util.Scanner;

public class String_Matching_1306_201308121604 {
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
while(sc.hasNext()){//8919883 2013-08-12 18:34:54 Accepted 1306	109MS	2848K	1342 B	Java	1983210400
String st = sc.next();
if(st.equalsIgnoreCase("-1"))
break;
String sr = sc.next();
int stlen = st.length();
int srlen = sr.length();

int count,max=0,i,j;
for( i=0; i<stlen; i++){
count=0;
int h=i;
for( j=0; j<srlen && h<stlen; ){//正面匹配
char st1 = st.charAt(h++);
char sr1 = sr.charAt(j++);
if(st1==sr1)
count++;
}
if(max<count)
max=count;

}

for( j=0; j<srlen; j++){
count=0;
int h=j;
for(i=0; i<stlen&& h<srlen; ){//反面匹配
char st1 = st.charAt(i++);
char sr1 = sr.charAt(h++);
if(st1==sr1)
count++;
}
if(max<count)
max=count;

}
int ss = stlen+srlen;
if(max==0)
System.out.println("appx"+"("+st+","+sr+") = "+0);
else{

if(ss%2*max==0){
int ss1 = ss/(2*max);
if(ss1!=1)
System.out.println("appx"+"("+st+","+sr+") = "+1+"/"+ss1);
else
System.out.println("appx"+"("+st+","+sr+") = "+1);
}

else{
System.out.println("appx"+"("+st+","+sr+") = "+2*max+"/"+ss);
}
}

}
}

}