浙大在线评测 1170 String Matching

王朝other·作者佚名  2006-01-09
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Problem:

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.

Input:

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.

Output:

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

Fractions reducing to zero or one should have no denominator.

Solution:

#include<iostream>

#include<string>

using namespace std;

// 声明:本代码仅供学习之用,请不要作为个人的成绩提交。

// http://blog.csdn.net/mskia

// email: bitrain@hotmail.com

int main( void ) {

for( ; ; ) {

string orig;

cin >> orig;

if ( orig == "-1" ) {

break;

}

string des;

cin >> des;

int lenA = orig.length( );

int lenB = des.length( );

int max = 0;

for ( int i = 0; i < lenA; ++i ) {

char strt = orig[ i ];

int copyi = i;

for( int z = 0; z < lenB; ++z ) {

int count = 0;

if ( strt == des[ z ] ) {

for ( int j = z; j < lenB; ++j ) {

if ( i == lenA ) {

break;

}

if ( orig[ i ] == des[ j ] ) {

++count;

}

++i;

}

}

if ( count > max ) {

max = count;

}

}

i = copyi;

}

for ( int i = 0; i < lenB; ++i ) {

char strt = des[ i ];

int copyi = i;

for ( int z = 0; z < lenA; ++z ) {

int count = 0;

if ( strt == orig[ z ] ) {

for ( int j = z; j < lenA; ++j ) {

if ( i == lenB ) {

break;

}

if ( orig[ i ] == des[ j ] ) {

++count;

}

++i;

}

}

if ( count > max ) {

max = count;

}

}

i = copyi;

}

if ( max == 0 ) {

cout << "appx(" << orig << "," << des << ") = " << 0 << endl;

} else if ( ( max + max ) % ( lenA + lenB ) == 0 ) {

cout << "appx(" << orig << "," << des << ") = " << (max + max) / (lenA + lenB) << endl;

} else {

int up = max + max, down = lenA + lenB, len = lenA + lenB;

for ( int i = 2; i < len; ++i ) {

if ( up % i == 0 && down % i == 0 ) {

up /= i;

down /= i;

i = 1;

}

}

cout << "appx(" << orig << "," << des << ") = " << up << "/" << down << endl;

}

}

return 0;

}

 
 
 
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