?? usaco 1_3_1 mixing milk 題解_leokan的blog.mht
字號(hào):
sorted[x]:=3Dt^.farmer;
inc(x);
t:=3Dt^.next;
end;
end;
sofar:=3D0;
for i:=3D1 to cnt do begin
if want < amount[sorted[i]] then begin
inc(sofar,want*cost[sorted[i]]);
want:=3D0; break;
end
else inc(sofar,amount[sorted[i]]*cost[sorted[i]]);
dec(want,amount[sorted[i]]);
end;
writeln(fOut,sofar);
close(fOut);
end.</PRE></DIV>
<P align=3Dleft><EM>Dwayne Crooks writes:</EM> Do we really need =
the linked=20
list that SVK's Adam Okruhlica uses in his solution, I don't think =
so.=20
Here is a better solution that uses essentially the same idea as =
Adam's=20
solution but does away with the linked list. Takes O(max(MAXP,M)) =
time,=20
where MAXP=3D1000 and M<=3D5000 is the input size. <EM>Editors =
note: Dwayne=20
should be using long long integers (64 bit) instead of int's in =
order to=20
avoid overflow.</EM></P>
<DIV align=3Dleft><PRE>#include <iostream>
#include <fstream>
#define MAXP 1000
using namespace std;
int main() {
ifstream in("milk.in");
ofstream out("milk.out");
=20
int N, M;
int P[MAXP+1];
=20
in >> N >> M;
for (int i =3D 0; i <=3D MAXP; i++) P[i]=3D0;
for (int i =3D 0; i < M; i++) {
int price, amt;
in >> price >> amt;
=20
// 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] +=3D amt;
}
=20
// greedy choice: take as much of the item that
// has the least price per gallon
int res =3D 0;
for (int p =3D 0; p<=3DMAXP && N>0; p++) {
if (P[p]>0) {
res+=3Dp*(N<P[p]?N:P[p]);
N-=3DP[p];
}
}
out << res << endl;
=20
in.close();
out.close();
=20
return 0;
}</PRE></DIV>
<P align=3Dleft><EM>As the final word, Bulgaria's Miroslav Paskov =
has=20
distilled all the best ideas into a simple solution:</EM></P>
<DIV align=3Dleft><PRE>#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 =3D 0;
fin>>needed>>farmers;
for(i =3D 0;i<farmers;i++){
fin>>price>>amount;
milk[price][0] +=3D amount; =20
}=20
for(i =3D 0; i<MAXPRICE && needed;i++){
if(needed> =3D milk[i][0]) {
needed -=3D milk[i][0];
paid +=3D milk[i][0] * i;
} else if(milk[i][0]>0) {
paid +=3D i*needed;
needed =3D 0; =20
}
}
fout << paid << endl;=20
return 0;
}</PRE></DIV></STRONG>
<P align=3Dleft></P>
<DIV align=3Dleft></DIV></DIV></TD></TR></TBODY></TABLE><BR>
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