UVa11054 Wine trading in Gergovia

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#uva #题解

本文解析了UVa11054葡萄酒交易问题，这是一个经典的贪心算法题目，与NOIP2002均分纸牌问题类似。文中提供了一个简洁的C++代码实现，用于最小化交易过程中酒的运输工作量。

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UVa11054 Wine trading in Gergovia

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

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 i-th 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’.

Output

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

Solution

就是一个最简单的贪心，跟noip2002均分纸牌很像

代码

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

typedef long long LL;
const int maxn = 100005;
LL ans, a[maxn];
int n;

int main()
{
    freopen("input.in", "r", stdin);
    freopen("output.out", "w", stdout);
    register int i;
    while(scanf("%d", &n) == 1) {
      if(n == 0) break;
      for(i = 1; i <= n; i++)
        scanf("%lld", &a[i]);
      ans = 0;
      for(i = 1; i < n; i++)
        if(a[i]) {
          ans += abs(a[i]);
          a[i + 1] += a[i];
        }
      printf("%lld\n", ans);
    }
    return 0;
}