我正在解决JAVA编程挑战中的鄂尔多斯数问题。 代码在我的机器中运行完美。但是在在线判断时会导致运行时错误。谁能指出我犯的错误?
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=985
这是代码
import java.util.*;
import java.io.*;
class Main
{
private String inputLines[];
private String namesToBeFound[];
private String namesInEachBook[][];
private String searchItem;
private boolean solnfound=false;
private static final BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
static String read() throws IOException
{
String line;
while(true)
{
line=br.readLine();
if(line==null) break; //eof
else if(line.length()==0) continue; //blank line
else
{
line=line.trim().replaceAll("\\s+"," ");
return line;
}
}
return null;
}
public static void main(String args[]) throws IOException
{
Main ob=new Main();
int totalPapers,calcAuthors,totalScenarios;
//First input number of scenarios
totalScenarios=Integer.parseInt(read());
//Now start a loop for reading total number of scenarios
for(int scenario=1;scenario<=totalScenarios;scenario++)
{
//Now read the line containing the number of papers and authors
StringTokenizer line=new StringTokenizer(read()," ");
totalPapers=Integer.parseInt(line.nextToken());
calcAuthors=Integer.parseInt(line.nextToken());
//Read a line containing author names along with book names
ob.inputLines=new String[totalPapers];
for(int i=0;i<totalPapers;i++)
ob.inputLines[i]=read();
//Read a line containing the names to be searched
ob.namesToBeFound=new String[calcAuthors];
for(int i=0;i<calcAuthors;i++)
ob.namesToBeFound[i]=read();
//Now generate the array
ob.buildArray();
//Now search
System.out.println("Scenario "+scenario);
for(int i=0;i<calcAuthors;i++)
{
ob.searchItem=ob.namesToBeFound[i];
if(ob.searchItem.equals("Erdos, P."))
{
System.out.println("Erdos, P. 0");
continue;
}
ob.search(ob.namesToBeFound[i],1,new ArrayList());
if(ob.solnfound==false) System.out.println(ob.searchItem+" infinity");
ob.solnfound=false;
}
}
}
private void buildArray()
{
String str;
namesInEachBook=new String[inputLines.length][];
for(int i=0;i<inputLines.length;i++)
{
str=inputLines[i];
str=str.substring(0,str.indexOf(':'));
str+=",";
namesInEachBook[i]=new String[(countCommas(str)+1)>>1];
for(int j=0;j<namesInEachBook[i].length;j++)
{
str=str.trim();
namesInEachBook[i][j]="";
namesInEachBook[i][j]+=str.substring(0,str.indexOf(','))+",";
str=str.substring(str.indexOf(',')+1);
namesInEachBook[i][j]+=str.substring(0,str.indexOf(','));
str=str.substring(str.indexOf(',')+1);
}
}
}
private int countCommas(String s)
{
int num=0;
for(int i=0;i<s.length();i++)
if(s.charAt(i)==',') num++;
return num;
}
private void search(String searchElem,int ernosDepth,ArrayList searchedElem)
{
ArrayList searchSpace=new ArrayList();
searchedElem.add(searchElem);
for(int i=0;i<namesInEachBook.length;i++)
for(int j=0;j<namesInEachBook[i].length;j++)
{
if(namesInEachBook[i][j].equals(searchElem)) //Add all authors name in this group
{
for(int k=0;k<namesInEachBook[i].length;k++)
{
if(namesInEachBook[i][k].equals("Erdos, P.")) //Found
{
solnfound=true;
System.out.println(searchItem+" "+ernosDepth);
return;
}
else if(searchedElem.contains(namesInEachBook[i][k]) || searchSpace.contains(namesInEachBook[i][k])) continue;
searchSpace.add(namesInEachBook[i][k]);
}
break;
}
}
Iterator i=searchSpace.iterator();
while(i.hasNext())
{
String cSearchElem=(String)i.next();
search(cSearchElem,ernosDepth+1,searchedElem);
}
}
}
答案 0 :(得分:1)
一个问题可能是:
totalScenarios=Integer.parseInt(read()),
如果输入的值不是NumberFormatException
,则可能会抛出int
。你需要使用try / catch来处理它。
答案 1 :(得分:1)
除了输入不包含NumberFormatException
时可能生成的int
之外,程序也不会以良好的方式处理输入终止。
您也不会使用任何记忆技术并且每次都建立相同的搜索树,这将导致UVa Judge出现Time Limit Exceeded
错误。
你还应该使用输入和放大器的缓冲。输出以进一步减少编译时间。减少对函数的调用,如果可能,可以内联它们,即在同一范围内写入。
希望这有帮助