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2013
11-09

POJ 1363 Rails [解题报告] Java

Rails

问题描述 :

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.



The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, ..., N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, ..., aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

输入:

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0.

The last block consists of just one line containing 0.

输出:

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last “null” block of the input.

样例输入:

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

样例输出:

Yes
No

Yes

解题代码:

方法一:
import java.util.*;
public class Main {
 
 public static void main(String[] args){

    Scanner sc = new Scanner(System.in);
    int in[]=new int[1001];   
    int out[]=new int[1001];   
    int stack[]=new int[1001];   

    int n;   
    int i,j;   
    for(i=0;i< 1001;i++)   
        in[i]=i+1;   
    while(sc.hasNext())   
    {   
         n=sc.nextInt();
         if(n==0) break;  
        while(true)   
        {   
            out[0]=sc.nextInt();
            if(out[0]==0)   
                break;   
            for(i=1;i< n;i++)   
                out[i]=sc.nextInt();   
            i=j=0;   
            int top=-1;   
            while(i< n)   
            {   
                top++;   
                stack[top]=in[i];   
                i++;   
                while(stack[top]==out[j])   
                {   
                    top--;   
                    j++;   
                    if(top==-1)break;   
                }   
            }   
            if(top==-1)   
                System.out.printf("Yes\n");   
            else  
                System.out.printf("No\n");   
        }   
        System.out.printf("\n");   
    }   
   }
} 


方法二:

//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;

public class Main {
 static final int N = 1000+10;
 static int n ;
 static int in[] = new int[N],out[] = new int[N],stack[] = new int[N];
 static boolean can(){
  int i=0,cnt=0,top=0;
  while(true){
	if(top>0 && stack[top-1]==out[cnt]){
		++cnt;
		--top;
		if(cnt>=n) break;
	}
	else{
		if(i>=n) return false;
		stack[top++] = in[i++];
	}
   }
   return true;
 }

  public static void main(String[]args) throws Exception{
  int i=0,j=0;
  StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
  while(true){
	n = Get_Num(cin);
	if(n==0) break;
	while(true){
		out[0] = Get_Num(cin);
		if(out[0]==0) break;
		for(i=1;i< n;++i) out[i] = Get_Num(cin);
		for(i=0;i< n;++i) in[i] = i+1;
		if(can()) System.out.println("Yes");
		else System.out.println("No");
	}
	System.out.println();
   }
 }
	static int Get_Num(StreamTokenizer cin) throws Exception{
		cin.nextToken();
		return (int)cin.nval;
	}
}

  1. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。