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2013
11-11

POJ 2457 Part Acquisition [解题报告] Java

Part Acquisition

问题描述 :

The cows have been sent on a mission through space to acquire a new milking machine for their barn. They are flying through a cluster of stars containing N (1 <= N <= 50,000) planets, each with a trading post.

The cows have determined which of K (1 <= K <= 1,000) types of objects (numbered 1..K) each planet in the cluster desires, and which products they have to trade. No planet has developed currency, so they work under the barter system: all trades consist of each party trading exactly one object (presumably of different types).

The cows start from Earth with a canister of high quality hay (item 1), and they desire a new milking machine (item K). Help them find the best way to make a series of trades at the planets in the cluster to get item K. If this task is impossible, output -1.

输入:

* Line 1: Two space-separated integers, N and K.

* Lines 2..N+1: Line i+1 contains two space-separated integers, a_i and b_i respectively, that are planet i’s trading trading products. The planet will give item b_i in order to receive item a_i.

输出:

* Line 1: One more than the minimum number of trades to get the milking machine which is item K (or -1 if the cows cannot obtain item K).

* Lines 2..T+1: The ordered list of the objects that the cows possess in the sequence of trades.

样例输入:

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

样例输出:

4
1
3
2
5

温馨提示:

OUTPUT DETAILS:

The cows possess 4 objects in total: first they trade object 1 for object 3, then object 3 for object 2, then object 2 for object 5.

解题代码:

//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;
public class Main {
	static final int N = 10000+10;
	static int meat[] = new int[N],pre[] = new int[N],n,K;
	static Vector Graph[] = new Vector[N];
	static Queue que = new LinkedList();
	static void start(){
		for(int i=0;i< N;++i)
			Graph[i] = new Vector();
	}
	static void init(){
		que.clear();
		for(int i=0;i< N;++i){
			Graph[i].clear();
			meat[i] = N+N;
			pre[i] = -1;
		}
		meat[1] = 1;
		que.offer(1);
	}

 public static void main(String[]args) throws Exception{
  int i,u,v;
  start();
  init();
  StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

  n = Get_Num(cin);
  K = Get_Num(cin);
  for(i=0;i< n;++i){
	v = Get_Num(cin);
	u = Get_Num(cin);
	Graph[v].add(u);
  }

  solve();
  if(meat[K]>=N) System.out.println(-1);
  else{
	System.out.println(meat[K]);
	make_ans();
  }
 }
 
  static void solve(){
   int i,j,k,temp;
   while(!que.isEmpty()){
	temp = (Integer) que.element();
	que.remove();
	k = Graph[temp].size();
	for(i=0;i< k;++i){
		j = (Integer) Graph[temp].elementAt(i);
		if(meat[j]>meat[temp]+1){
			meat[j] = meat[temp]+1;
			pre[j] = temp;
			que.add(j);
		}
	}
   }
 }

  static void make_ans(){
   int ans[] = new int[N],i,k=K,top=0;
   while(true){
	ans[top++] = k;
	k = pre[k];
	if(k< 0) break;
   }
   PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
   for(i=top-1;i>=0;--i)
	out.println(ans[i]);
   out.flush();
 }

  static int Get_Num(StreamTokenizer cin) throws Exception{
   cin.nextToken();
   return (int) cin.nval;
 }
}

  1. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count

  2. 您没有考虑 树的根节点是负数的情况, 若树的根节点是个很大的负数,那么就要考虑过不过另外一边子树了