UVa 11054/HDU 1489/POJ 2940 Wine trading in Gergovia(贪心&双向队列)

11054 - Wine trading in Gergovia

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=113&page=show_problem&problem=1995

http://acm.hdu.edu.cn/showproblem.php?pid=1489

http://poj.org/problem?id=2940

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.

Input Specification

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 a_i (-1000 ≤ a_i ≤ 1000). If a_i ≥ 0, it means that the inhabitant living in the i^th house wants to buy a_ibottles of wine, otherwise if a_i < 0, he wants to sell -a_i bottles of wine. You may assume that the numbers a_i sum up to 0.
The last test case is followed by a line containing 0.

Output Specification

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", in JAVA the data type "long").

Sample Input

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

Sample Output

9
9000

思路：

一开始我想的是用双向队列存储，模仿：1. 顾客排队和售货商的排队 2. 回合制游戏

(注意，就算正数是卖出，负数是买入，结果也是一样的。)

但是，后来我发现不用这么做，

一种新思路：

假设有一辆马车，从左往右行驶，它可以代替人们买卖葡萄酒！你可能会问，如果一开始遇到的都是想买的人呢？可以这么想，从买家到卖家和从卖家到买家是等价的。所以整个程序可以顺序一遍搞定。

完整代码：

双向队列的，慢死了：

PS：在UVaOJ上请使用%lld替代%I64d

/*UVaOJ: 0.065s*/
/*HDU: 62ms,1740KB*/
/*POJ: 375ms,2056KB*/

#include <cstdio>
#include <deque>
#include <algorithm>
using namespace std;

struct node
{
	int pos, num;
} a[100000];

deque<node> buy, sell;

int main()
{
	int n, temp;
	while (scanf("%d", &n), n)
	{
		buy.clear();
		sell.clear();
		node tempbuy, tempsell;
		long long sum = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &temp);
			a[i].pos = i;
			a[i].num = temp;
			if (temp > 0)
				buy.push_back(a[i]);
			else if (temp < 0)
				sell.push_back(a[i]);//注意a[i].num是负数

			//while里面的操作就像回合制游戏一样，你懂的~
			while (!(buy.empty() || sell.empty()))
			{
				tempbuy = buy.front();
				buy.pop_front();
				tempsell = sell.front();
				sell.pop_front();
				if (tempbuy.num + tempsell.num > 0)
				{
					sum -= tempsell.num * abs(tempbuy.pos - tempsell.pos);
					tempbuy.num += tempsell.num;
					buy.push_front(tempbuy);
				}
				else
				{
					if (tempbuy.num + tempsell.num < 0)
					{
						tempsell.num += tempbuy.num;
						sell.push_front(tempsell);
					}
					sum += tempbuy.num * abs(tempbuy.pos - tempsell.pos);
				}
			}
		}
		printf("%I64d\n",sum);
	}
	return 0;
}

新思路：

PS：在UVaOJ上请使用%lld替代%I64d

/*UVaOJ: 0.048s*/
/*HDU: 31ms,228KB*/
/*POJ: 94ms,144KB*/

#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
	int n, a;
	long long temp, ans;
	while (scanf("%d", &n), n)
	{
		temp = ans = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &a);
			temp += a;
            ans += abs(temp);
		}
		printf("%I64d\n", ans);
	}
	return 0;
}