HDU-1306-String Matching
来源:互联网 发布:苹果电脑ps软件 编辑:程序博客网 时间:2024/06/05 18:27
String Matching
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 776 Accepted Submission(s): 404
Problem 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:
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.
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.
Sample 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 CARTTURKEY CHICKENMONEY POVERTYROUGH PESKYA A-1The words will all be uppercase.
Sample Output
Print the value for appx() for each pair as a reduced fraction, like this:appx(CAR,CART) = 6/7appx(TURKEY,CHICKEN) = 4/13appx(MONEY,POVERTY) = 1/3appx(ROUGH,PESKY) = 0appx(A,A) = 1
Recommend
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 1306109MS2848K1342 BJava1983210400String 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);elseSystem.out.println("appx"+"("+st+","+sr+") = "+1);}else{System.out.println("appx"+"("+st+","+sr+") = "+2*max+"/"+ss);}}}} }
- hdu-1306- String Matching
- HDU-1306-String Matching
- hdu 1306 String Matching
- hdu 1036 String Matching
- HDU 3407 String-Matching Automata
- POJ 1580 && HDU 1306 String Matching(水~)
- hdoj 1306 String Matching
- hdu 3259 Just Another String Matching Problem
- 杭电1306 String Matching
- KMP Matching: String Matching
- String Matching
- String Matching
- String Matching
- Naive String Matching: String Matching
- HDU--3407[String-Matching Automata] AC自动机或kmp
- 杭电acm—1306 String Matching
- Finite Automaton Matching: String Matching
- Rabin Karp Matching: String Matching
- zoj 3103 Cliff Climbing (SPFA)
- n!后面有多少个0
- linux grep命令
- oracle union 用法
- SQLSERVER2005 跨数据库访问 -- 建立链接服务器 .
- HDU-1306-String Matching
- rabbitMQ小小的入门
- p2p 了解资源整合
- POJ 3422 Kaka's Matrix Travels 解题报告(最大费用最大流)
- 通过反射调用Android的L2CAP接口
- hdu 1847 Good Luck in CET-4 Everybody!
- PHP 多线程的实现 curl_multi
- 嵌入式加密芯片dm2016驱动
- Android Bluetooth蓝牙开发\蓝牙协议\蓝牙通信例子_Android支持蓝牙4.0版本_BLE开发