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2013
11-09

POJ 1511 Invitation Cards [解题报告] Java

Invitation Cards

问题描述 :

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where ‘X’ denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

输入:

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

输出:

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

样例输入:

2
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50

样例输出:

46
210

解题代码:

//* @author: [email protected]
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main
{
 static int[] cost;
 static node[] edge,redge;
 static int n;
 static boolean[] used;
 public static void main(String[] args) throws NumberFormatException, IOException
  {
   InputStreamReader is=new InputStreamReader(System.in);
   BufferedReader in=new BufferedReader(is);
   int h=Integer.parseInt(in.readLine());
   while((h--)!=0)
   {
	String[] ss=in.readLine().split(" ");
	n=Integer.parseInt(ss[0]);
	edge=new node[n];
	redge=new node[n];
	cost=new int[n];
	used=new boolean[n];
	for(int i=0;i< n;i++)
	{
		edge[i]=new node(-1,-1);
		redge[i]=new node(-1,-1);
	}
	int m=Integer.parseInt(ss[1]);
	for(int i=0;i< m;i++)
	{
		ss=in.readLine().split(" ");
		int x=Integer.parseInt(ss[0]);
		int y=Integer.parseInt(ss[1]);
		int w=Integer.parseInt(ss[2]);
		insert(x,y,w);
	}
	long w=0;
	w+=spfa(0);
	w+=spfa(1);
	System.out.println(w);
   }
  }

  static void insert(int x,int y,int w)
  {
    x--;y--;
    node temp=new node(y,w);
    temp.next=edge[x].next;
    edge[x].next=temp;
    temp=new node(x,w);
    temp.next=redge[y].next;
    redge[y].next=temp;	
   }

  static long spfa(int d)
  {
   node temp;
   for(int i=0;i< n;i++)
	cost[i]=2000000000;
   Queue< Integer> qu=new LinkedList< Integer>();
   qu.add(0);
   cost[0]=0;
   while(!qu.isEmpty())
   {
	int u=qu.poll();
	used[u]=false;
	if(d==0)temp=edge[u].next;
	else temp=redge[u].next;
	while(temp!=null)
	{
		if(cost[temp.v]>cost[u]+temp.w)
		{
		  cost[temp.v]=cost[u]+temp.w;
		  if(!used[temp.v])
		  {
			qu.add(temp.v);
			used[temp.v]=true;
		  }
		}
		
		temp=temp.next;
	}
   }
  long ans=0;
  for(int i=0;i< n;i++)
	ans+=cost[i];
  return ans;
 }
	
}

class node
{
	int v,w;
	node next=null;
	public node(int vv,int ww)
	{
		v=vv;w=ww;
	}
}

  1. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  2. 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。

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  4. 第二块代码if(it != mp.end())应改为if(it != mp.end() && (i+1)!=(it->second +1));因为第二种解法如果数组有重复元素 就不正确

  5. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同,其树的前、中、后序遍历是相同的,但在此处不能使用中序遍历,因为,中序遍历的结果就是排序的结果。经在九度测试,运行时间90ms,比楼主的要快。

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