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[POJ] String Matching
String Matching
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 4074 | Accepted: 2077 |
Description
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:
with two common letters (A and R), but the best is:
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.
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:
-----------------------------
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.
The words will all be uppercase.
Using the above technique, you are to calculate appx() for the pair of words on the line and print the result.
The words will all be uppercase.
Output
Print the value for appx() for each pair as a reduced fraction,Fractions reducing to zero or one should have no denominator.
Sample Input
CAR CART TURKEY CHICKEN MONEY POVERTY ROUGH PESKY A A -1
Sample Output
appx(CAR,CART) = 6/7 appx(TURKEY,CHICKEN) = 4/13 appx(MONEY,POVERTY) = 1/3 appx(ROUGH,PESKY) = 0 appx(A,A) = 1
字符匹配问题:
按着题目要求写就好了
记住要相反方向进行两次求值
#include<iostream> #include<string> using namespace std; int appx(string& word1,string&word2) { int count = 0; int max = 0; int length1 = word1.length(); int length2 = word2.length(); for (int i = 0; i<length1; i++) { count = 0; for (int j = 0; j<length2&&i + j<length1; j++) { if (word1[i + j] == word2[j]) count++; } if (max<count) max=count; } return max; } int main() { string word1; string word2; while (cin >> word1&&word1 != "-1") { cin >> word2; int len1 = word1.length(); int len2 = word2.length(); int app1 = appx(word1,word2); int app2 = appx(word2,word1); if (app1<app2)app1 = app2; cout << "appx("; for (int i = 0; i<len1; i++) cout << word1[i]; cout << ","; for (int i = 0; i<len2; i++) cout << word2[i]; cout << ") = "; if (app1 == 0)cout << 0 << endl; else { len1 += len2; app1 *= 2; for (int i = 2; i <= ((len1<app1) ? len1 : app1); i++) while (app1%i == 0 && len1%i == 0) { app1 /= i; len1 /= i; } if (app1%len1 != 0) cout << app1 << ‘/‘ << len1 << endl; else cout << app1 / len1 << endl; } } return 0; }
[POJ] String Matching
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