首页 > 专题系列 > Java解POJ > POJ 2940 Wine Trading in Gergovia [解题报告] Java
2013
11-12

POJ 2940 Wine Trading in Gergovia [解题报告] Java

Wine Trading in Gergovia

问题描述 :

As you may know from the comic “Asterix and the Chieftain’s Shield”, Gergovia consists of one street, and every inhabitant of the city is a wine salesman. You wonder how this economy works? Simple enough: everyone buys wine from other inhabitants of the city. Every day each inhabitant decides how much wine he wants to buy or sell. Interestingly, demand and supply is always the same, so that each inhabitant gets what he wants.

There is one problem, however: Transporting wine from one house to another results in work. Since all wines are equally good, the inhabitants of Gergovia don’t care which persons they are doing trade with, they are only interested in selling or buying a specific amount of wine. They are clever enough to figure out a way of trading so that the overall amount of work needed for transports is minimized.

In this problem you are asked to reconstruct the trading during one day in Gergovia. For simplicity we will assume that the houses are built along a straight line with equal distance between adjacent houses. Transporting one bottle of wine from one house to an adjacent house results in one unit of work.

输入:

The input consists of several test cases.

Each test case starts with the number of inhabitants n (2 ≤ n ≤ 100000). The following line contains n integers ai (−1000 ≤ ai ≤ 1000). If ai ≥ 0, it means that the inhabitant living in the ith house wants to buy ai bottles of wine, otherwise if ai < 0, he wants to sell −ai bottles of wine. You may assume that the numbers ai sum up to 0.

The last test case is followed by a line containing 0.

输出:

For each test case print the minimum amount of work units needed so that every inhabitant has his demand fulfilled. You may assume that this number fits into a signed 64-bit integer (in C/C++ you can use the data type “long long” or “__int64”, in JAVA the data type “long”).

样例输入:

5
5 -4 1 -3 1
6
-1000 -1000 -1000 1000 1000 1000
0

样例输出:

9
9000

解题代码:

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

public class Main {
	
	static final int N = 100000+100;
	static int n;
	static long num[] = new long[N];
	
	public static double Get_Num(StreamTokenizer cin) throws Exception{
		cin.nextToken();
		return cin.nval;
	}
	
public static void main(String []args) throws Exception{
		
 int i;
 StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
		
 while(true){
	n = (int) Get_Num(cin);
	if(n==0) break;
	for(i=1;i<=n;++i)
		num[i] = (long) Get_Num(cin);
	System.out.println(solve());
  }
}

public static long solve(){
		int i;
		long cnt=0,ans=0;
		for(i=1;i<=n;++i){
			cnt +=num[i];
			ans+=abs(cnt);
		}
		return ans;
	}
	public static long abs(long cnt){
		if(cnt< 0) return -cnt;
		return cnt;
	}
}

  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. 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])