UVa 11054：Wine trading in Gergovia

【連結】

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

【原題】

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 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.

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

55 -4 1 -3 16-1000 -1000 -1000 1000 1000 10000

Sample Output

99000

【題目大意】

一條街上住著連續的n戶人家，沒相鄰的兩戶人相隔一個單位。街上的每戶人都需要買一定數量的葡萄酒或者賣掉葡萄酒，保證所有人家買進的總量與賣出的數量一致。每戶可以選擇與其他任何家交易。但是因為相隔路程不一樣，所以需要路費。路費是按照交易量*相隔距離算的。求所有人都交易滿足，最小的路費總和是多少。

【分析與總結】

本欄目更多精彩內容：http://www.bianceng.cn/Programming/sjjg/

由於路費和距離相關，所以需要讓路程越小越好，那麼就可以讓每人之和相鄰的人交易。 不用考慮他們是要買還是要賣。假設a1要買5個，a2要賣2個，那麼就讓a1向a2買5個，不用管a2有多少，不夠可以打欠條，就變成-3了，這時a3就要向a3買3個。如此算下去，只需要從第一個枚舉到最後一個便可得到最小的答案。

【代碼】

/*  *  UVa: 11054 - Wine trading in Gergovia   *  Result: Accept  *  Time: 0.044s  *  Author: D_Double  */#include<iostream>  #include<cstdio>  #include<cstring>  #include<cmath>  using namespace std;        int arr[100002];        int main(){      int n;      while(scanf("%d",&n) && n){          for(int i=0; i<n; ++i)              scanf("%d",&arr[i]);          long long ans=0;          for(int i=0; i<n-1; ++i){              ans += abs(arr[i]);              arr[i+1] += arr[i];          }          printf("%lld\n", ans);      }      return 0;  }