首页 > 专题系列 > Java解POJ > POJ 1598 Excuses, Excuses! [解题报告] Java
2013
11-10

POJ 1598 Excuses, Excuses! [解题报告] Java

Excuses, Excuses!

问题描述 :

Judge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time required listening to goofy excuses, Judge Ito has asked that you write a program that will search for a list of keywords in a list of excuses identifying lame excuses. Keywords can be matched in an excuse regardless of case.

输入:

Input to your program will consist of multiple sets of data. Line 1 of each set will contain exactly two integers. The first number (1 <= K <= 20) defines the number of keywords to be used in the search. The second number (1 <= E <= 20) defines the number of excuses in the set to be searched. Lines 2 through K+1 each contain exactly one keyword. Lines K+2 through K+1+E each contain exactly one excuse. All keywords in the keyword list will contain only contiguous lower case alphabetic characters of length L (1 <= L <= 20) and will occupy columns 1 through L in the input line. All excuses can contain any upper or lower case alphanumeric character, a space, or any of the following punctuation marks [".,!?] not including the square brackets and will not exceed 70 characters in length. Excuses will contain at least 1 non-space character.

输出:

For each input set, you are to print the worst excuse(s) from the list. The worst excuse(s) is/are defined as the excuse(s) which contains the largest number of incidences of keywords. If a keyword occurs more than once in an excuse, each occurrance is considered a separate incidence. A keyword “occurs” in an excuse if and only if it exists in the string in contiguous form and is delimited by the beginning or end of the line or any non-alphabetic character or a space.

For each set of input, you are to print a single line with the number of the set immediately after the string “Excuse Set #”. (See the Sample Output). The following line(s) is/are to contain the worst excuse(s) one per line exactly as read in. If there is more than one worst excuse, you may print them in any order. After each set of output, you should print a blank line.

样例输入:

5 3
dog
ate
homework
canary
died
My dog ate my homework.
Can you believe my dog died after eating my canary... AND MY HOMEWORK?
This excuse is so good that it contain 0 keywords.
6 5
superhighway
crazy
thermonuclear
bedroom
war
building
I am having a superhighway built in my bedroom.
I am actually crazy.
1234567890.....,,,,,0987654321?????!!!!!!
There was a thermonuclear war!
I ate my dog, my canary, and my homework ... note outdated keywords?

样例输出:

Excuse Set #1
Can you believe my dog died after eating my canary... AND MY HOMEWORK?

Excuse Set #2
I am having a superhighway built in my bedroom.
There was a thermonuclear war!

解题代码:

//* @author popop0p0popo
import java.util.*;
import java.io.*;

public class Main{
 public static void main(String[] args){
  Scanner scanner=new Scanner(new BufferedReader(new InputStreamReader(System.in)));
	String[] kws;
	String[] exs,exsk;
	int[] kn;
	String fl;
	int n,m,max;
	int index=1;
	while (scanner.hasNext()){
    	fl=scanner.nextLine();
		String[] t=fl.split(" ");
		n=Integer.parseInt(t[0]);
		m=Integer.parseInt(t[1]);
		kws=new String[n];
		exsk=new String[m];
		exs=new String[m];
		kn=new int[m];
		for (int i=0;i < n ;i++ ){
			kws[i]=scanner.nextLine().toLowerCase();
		}
	for (int i=0;i< m ;i++ ){
		exsk[i]=scanner.nextLine();
		exs[i]=exsk[i].toLowerCase();
		String tmp="";
		for (int j=0;j< exs[i].length() ;j++ ){
		if (Character.isLetter(exs[i].charAt(j))){
			tmp=tmp+exs[i].charAt(j);
		}
		else{
			tmp=tmp+"#";
		}
	}
	exs[i]=tmp;
	for (int j=0;j< n ;j++ ){
		if (exs[i].indexOf("#"+kws[j]+"#")!=-1){
			kn[i]++;
		}
		else if (exs[i].indexOf(kws[j]+"#")==0){
			kn[i]++;
		}
    else if (exs[i].indexOf("#"+kws[j])+kws[j].length()+1==exs[i].length()){
			kn[i]++;
		}
	}
 }
	max=kn[0];
	for (int i=1;i< kn.length ;i++ ){
		if (kn[i]>max){
			max=kn[i];
	}
}
System.out.println("Excuse Set #"+index++);
  for (int i=0;i< m ;i++ ){
	if (kn[i]==max){
		System.out.println(exsk[i]);
	}
  }
			System.out.println();
		}
	}
}

  1. 有限自动机在ACM中是必须掌握的算法,实际上在面试当中几乎不可能让你单独的去实现这个算法,如果有题目要用到有限自动机来降低时间复杂度,那么这种面试题应该属于很难的级别了。

  2. 我还有个问题想请教一下,就是感觉对于新手来说,递归理解起来有些困难,不知有没有什么好的方法或者什么好的建议?

  3. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。