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.
PROGRAM NAME: milk
INPUT FORMAT
| Line 1: | Two integers, N and M. The first value, N, (0 <= N <= 2,000,000) is the amount of milk that Merry Milk Makers wants per day. The second, M, (0 <= M <= 5,000) is the number of farmers that they may buy from. |
| Lines 2 through M+1: | The next M lines 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. |
SAMPLE INPUT (file milk.in)
100 5 5 20 9 40 3 10 8 80 6 30
OUTPUT FORMAT
A single line with a single integer that is the minimum price that Merry Milk Makers can get their milk at for one day.
SAMPLE OUTPUT (file milk.out)
630
这题是我在奶牛网上A的最简单的题了，不解释。
/*
ID:nealgav1
PROG:milk
LANG:C++
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 1234567
using namespace std;
class milk
{
public:
int per;
int volume;
};
milk cow[N];
bool cmp(milk a,milk b)
{
return a.per<b.per;
}
int main()
{
freopen("milk.in","r",stdin);
freopen("milk.out","w",stdout);
int require,num;
while(scanf("%d%d",&require,&num)!=EOF)
{
for(int i=0;i<num;i++)
{
scanf("%d%d",&cow[i].per,&cow[i].volume);
}
sort(cow,cow+num,cmp);
int ans=0;
int j=0;
while(require)
{
if(cow[j].volume>=require)
{
ans+=require*cow[j].per;
require-=require;
}
else
{
ans+=cow[j].volume*cow[j].per;
require-=cow[j].volume;
j++;
}
}
printf("%d\n",ans);
}
}
Russ Cox
Since we're acquiring things that are all of the same size (in this case, units of milk), a greedy solution will suffice: we sort the farmers by price, and then buy milk from the farmers with the lowest prices, always completely exhausting one farmer's supply before moving on to the next one.
To do this, we read the input into Farmer structures, sort the array by price, and then walk the array, buying milk until we've got all the milk we want.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#define MAXFARMER 5000
typedef struct Farmer Farmer;
struct Farmer {
int p; /* price per gallon */
int a; /* amount to sell */
};
int
farmcmp(const void *va, const void *vb)
{
return ((Farmer*)va)->p - ((Farmer*)vb)->p;
}
int nfarmer;
Farmer farmer[MAXFARMER];
void
main(void)
{
FILE *fin, *fout;
int i, n, a, p;
fin = fopen("milk.in", "r");
fout = fopen("milk.out", "w");
assert(fin != NULL && fout != NULL);
fscanf(fin, "%d %d", &n, &nfarmer);
for(i=0; i<nfarmer; i++)
fscanf(fin, "%d %d", &farmer[i].p, &farmer[i].a);
qsort(farmer, nfarmer, sizeof(farmer[0]), farmcmp);
p = 0;
for(i=0; i<nfarmer && n > 0; i++) {
/* take as much as possible from farmer[i], up to amount n */
a = farmer[i].a;
if(a > n)
a = n;
p += a*farmer[i].p;
n -= a;
}
fprintf(fout, "%d\n", p);
exit(0);
}
Ran Pang of Canada writes:
Here is a program that solves the problem in linear time (with respect to the maximum price, and number of farmers, since we have to read in the data anyway), while I think the qsort used by the solution would consume O(n log n), where n is the number of farmers.
#include<stdio.h>
#define MAXPRICE 1001
int amount_for_price[MAXPRICE]={0};
int N, M;
int Cal(void);
int Read(void);
int main(void) {
Read();
Cal();
return 0;
}
int Cal(void) {
int i;
int price_total=0;
int milk_total=0;
for(i=0;i<MAXPRICE;i++) {
if(amount_for_price[i]) {
if(milk_total+amount_for_price[i]<N) {
price_total+=(i*amount_for_price[i]);
milk_total+=amount_for_price[i];
}
else {
int amount_needed = N-milk_total;
price_total+=(i*amount_needed);
break;
}
}
}
{
FILE* out=fopen("milk.out","w");
fprintf(out,"%d\n",price_total);
fclose(out);
}
return 0;
}
int Read(void) {
FILE* in = fopen("milk.in","r");
int i, price, amount;
fscanf(in,"%d %d",&N,&M);
for(i=0;i<M;i++) {
fscanf(in, "%d %d", &(price), &(amount));
amount_for_price[price]+=amount;
}
fclose(in);
return 0;
}
\f2Here is another solution from SVK's Adam Okruhlica\fP
It is unnecessary to sort the prices with quicksort in O(n.lg.n) time, because there is an upper limit of the a single price ($1000) and we know that all prices are integral. We can sort this array with count sort. We establish a 'box' for each of the available prices (0..1000). We save the input to an array. Then we iterate through each farmer and we memoize his index in the (0..1000) array on index equivalent to the price he offers us. Hence there can be more farmers offering the same price we put them in a linked list. Finally we iterate the array from 0 to 1000 andpick the farmers' indexes from the linked lists. It's pretty easy to implement, and the time complexity is O(n).
program milk;
type pList = ^List;
List = record
farmer:longint;
next:pList;
end;
HeadList = record
head:pList;
tail:pList;
end;
var fIn,fOut:text;
sofar,i,x,want,cnt,a,b:longint;
sorted,cost,amount:array[1..5010] of longint;
csort:array[0..1010] of HeadList;
t:pList;
begin
assign(fIn,'milk.in');reset(fIn);
assign(fOut,'milk.out'); rewrite(fOut);
readln(fIn,want,cnt);
for i:=1 to cnt do readln(fIn,cost[i],amount[i]);
for i:=0 to 1000 do begin
new(csort[i].head);
csort[i].tail:=csort[i].head;
csort[i].head^.farmer:=-1;
end;
{Cast indexes into the array}
for i:=1 to cnt do begin
t:=csort[cost[i]].tail;
if t^.farmer = -1 then t^.farmer:=i;
new(t^.next);
t^.next^.farmer:=-1;
csort[cost[i]].tail:=t^.next;
end;
{Pick indexes}
x:=1;
for i:=0 to 1000 do begin
t:=csort[i].head;
while t^.farmer > 0 do begin
sorted[x]:=t^.farmer;
inc(x);
t:=t^.next;
end;
end;
sofar:=0;
for i:=1 to cnt do begin
if want < amount[sorted[i]] then begin
inc(sofar,want*cost[sorted[i]]);
want:=0; break;
end
else inc(sofar,amount[sorted[i]]*cost[sorted[i]]);
dec(want,amount[sorted[i]]);
end;
writeln(fOut,sofar);
close(fOut);
end.
Dwayne Crooks writes: Do we really need the linked list that SVK's Adam Okruhlica uses in his solution, I don't think so. Here is a better solution that uses essentially the same idea as Adam's solution but does away with the linked list. Takes O(max(MAXP,M)) time, where MAXP=1000 and M<=5000 is the input size. Editors note: Dwayne should be using long long integers (64 bit) instead of int's in order to avoid overflow.
#include <iostream>
#include <fstream>
#define MAXP 1000
using namespace std;
int main() {
ifstream in("milk.in");
ofstream out("milk.out");
int N, M;
int P[MAXP+1];
in >> N >> M;
for (int i = 0; i <= MAXP; i++) P[i]=0;
for (int i = 0; i < M; i++) {
int price, amt;
in >> price >> amt;
// we can add amounts that cost the same price
// since x gallons costing c cents and
// y gollons costing c cents
// is the same as
// x+y gallons costing c cents
P[price] += amt;
}
// greedy choice: take as much of the item that
// has the least price per gallon
int res = 0;
for (int p = 0; p<=MAXP && N>0; p++) {
if (P[p]>0) {
res+=p*(N<P[p]?N:P[p]);
N-=P[p];
}
}
out << res << endl;
in.close();
out.close();
return 0;
}
As the final word, Bulgaria's Miroslav Paskov has distilled all the best ideas into a simple solution:
#include <fstream>
#define MAXPRICE 1001
using namespace std;
int main() {
ifstream fin ("milk.in");
ofstream fout ("milk.out");
unsigned int i, needed, price, paid, farmers, amount, milk[MAXPRICE][2];
paid = 0;
fin>>needed>>farmers;
for(i = 0;i<farmers;i++){
fin>>price>>amount;
milk[price][0] += amount;
}
for(i = 0; i<MAXPRICE && needed;i++){
if(needed> = milk[i][0]) {
needed -= milk[i][0];
paid += milk[i][0] * i;
} else if(milk[i][0]>0) {
paid += i*needed;
needed = 0;
}
}
fout << paid << endl;
return 0;
}
