标签：ack seve from += price alc 就是 个人 important
Since milk packaging is such a low margin business, it is important to keep the price of the raw product (milk) as low as possible. Help Merry Milk Makers get the milk they need in the cheapest possible manner.
The Merry Milk Makers company has several farmers from which they may buy milk, and each one has a (potentially) different price at which they sell to the milk packing plant. Moreover, as a cow can only produce so much milk a day, the farmers only have so much milk to sell per day. Each day, Merry Milk Makers can purchase an integral amount of milk from each farmer, less than or equal to the farmer‘s limit.
Given the Merry Milk Makers‘ daily requirement of milk, along with the cost per gallon and amount of available milk for each farmer, calculate the minimum amount of money that it takes to fulfill the Merry Milk Makers‘ requirements.
Note: The total milk produced per day by the farmers will be sufficient to meet the demands of the Merry Milk Makers.
The first line contains two integers, N and M. The first value, N, (0 <= N <= 2,000,000) is the amount of milk that Merry Milk Makers‘ want per day. The second, M, (0 <= M <= 5,000) is the number of farmers that they may buy from.
The next M lines (Line 2 through M+1) each contain two integers, Pi and Ai. Pi (0 <= Pi <= 1,000) is price in cents that farmer i charges. Ai (0 <= Ai <= 2,000,000) is the amount of milk that farmer i can sell to Merry Milk Makers per day.
A single line with a single integer that is the minimum price that Merry Milk Makers can get their milk at for one day.
100 5 5 20 9 40 3 10 8 80 6 30
630
#include<iostream> #define FARMERS 5000 using namespace std; int findMin(int *a,int n,bool *flag){ int min=1001; int count; for(int i=0;i<n;i++){ if(flag[i]==true)continue; if(a[i]<min){ min=a[i]; count=i; } } flag[count]=true; return count; } int main(){ bool flag[FARMERS]={false}; int all,farmers; cin>>all>>farmers; int price[FARMERS],amount[FARMERS]; int buy=0; for(int i=0;i<farmers;i++){ cin>>price[i]>>amount[i]; } while(all>0){ int min=findMin(price,farmers,flag); int this_buy=amount[min]; if(this_buy>all) this_buy=all; all-=this_buy; buy+=this_buy*price[min]; } cout<<buy<<endl; return 0; }
注：flag数组的作用是打标记。因为找到最便宜的一家之后，再找第二便宜的一家 正常的思路是找到最便宜的，买完后，删掉它。再再数组里找最小的，重复前面操作。但是考虑到删除数组中的一个元素是很麻烦的，所以采取了打标记的方式，其实就是把它当做已经删除了~
findMin这个函数返回的是当前最便宜的那家的下标。变量all最初是他要买的牛奶总数，随着while循环不断迭代，all记录的就是还需要买的牛奶数量，直到它为0，就代表买完了。
标签：ack seve from += price alc 就是 个人 important
原文地址：https://www.cnblogs.com/qq1337822982/p/10891655.html
