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2013
11-12

POJ 3259 Wormholes [解题报告] Java

Wormholes

问题描述 :

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

输入:

Line 1: A single integer, F. F farm descriptions follow.

Line 1 of each farm: Three space-separated integers respectively: N, M, and W

Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.

Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

输出:

Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

样例输入:

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

样例输出:

NO
YES

温馨提示:

For farm 1, FJ cannot travel back in time.

For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

解题代码:

//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;
class edge{
	int s,t,l;
	public void set(int _s,int _t,int _l){
		this.s = _s;
		this.t = _t;
		this.l = _l;
	}
}
public class Main {
	static int M = 6000,N = 500+20,Inf = 1000000000+10;
	static edge e[] =  new edge[M];
	static void start(){
		for(int i=0;i< M;++i)
			e[i] = new edge();
	}
	
static boolean Bellman_Ford(int n,int m,int d[],int s,edge e[]){
  int i,j;
  boolean flag;
  for(i=0;i<=n;++i) d[i] = Inf;
  d[s] = 0;
  for(i=0;i<=n;++i){
	flag = false;
	for(j=0;j< m;++j){
		if(d[e[j].t]>d[e[j].s]+e[j].l){
			d[e[j].t] = d[e[j].s]+e[j].l;
			flag = true;
		}
	}
	if(!flag) break;
   }
   for(i=0;i< m;++i)
	if(d[e[i].t]>d[e[i].s]+e[i].l) return false;
   return true;
}
	
public static void main(String[]args) throws Exception{
  int d[] = new int[N];
  int n,m,w,i,s,t,l,top,F;
  start();
  StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		
  F = Get_Num(cin);
  while(F--!=0){
	top = 0;
	n = Get_Num(cin);
	m = Get_Num(cin);
	w = Get_Num(cin);
	for(i=0;i< m;++i){
		s = Get_Num(cin);
		t = Get_Num(cin);
		l = Get_Num(cin);
		
		e[top++].set(s, t, l);
		e[top++].set(t, s, l);
	}
	for(i=0;i< w;++i){
		s = Get_Num(cin);
		t = Get_Num(cin);
		l = Get_Num(cin);
		e[top++].set(s, t, -l);
	}
	for(i=1;i<=n;++i){
		e[top++].set(0,i,0);
	}
	if(Bellman_Ford(n,top,d,0,e))
		System.out.println("NO");
	else System.out.println("YES");
   }
		
  }
	static int Get_Num(StreamTokenizer cin)throws Exception{
		cin.nextToken();
		return (int)cin.nval;
	}
}

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。

  3. 第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。