?? usaco 3_2_4 feed ratios 題解_leokan的blog.mht
字號:
exit(num);<BR>end;<BR>procedure work;<BR>var<BR> =
d,i,j,k:longint;<BR> dd:array[1..3] of=20
longint;<BR> =
temp:matrix;<BR>begin<BR> =20
assign(output,'ratios.out');rewrite(output);<BR> =
d:=3Ddeterminant(feed);<BR> if d=3D0=20
then<BR> =20
=
begin<BR> &nbs=
p;=20
=
writeln('NONE');<BR>  =
; =20
=
close(output);<BR> &=
nbsp; =20
exit;<BR> =20
end;<BR> for i:=3D1 to 100=20
do<BR> =20
=
begin<BR> &nbs=
p;=20
for j:=3D1 to 3=20
=
do<BR> &=
nbsp; =20
=
begin<BR> &nbs=
p; =20
=
temp:=3Dfeed;<BR> &n=
bsp; =20
for k:=3D1 to 3=20
=
do<BR> &=
nbsp; =20
=
temp[k,j]:=3Dgoal[k]*i;<BR> &nbs=
p; =20
=
dd[j]:=3Ddeterminant(temp);<BR> =
=20
=
end;<BR>  =
;=20
if dd[1] mod d<>0 then=20
=
continue;<BR> =
=20
if dd[2] mod d<>0 then=20
=
continue;<BR> =
=20
if dd[3] mod d<>0 then=20
=
continue;<BR> =
=20
if dd[1] div d<0 then=20
=
continue;<BR> =
=20
if dd[2] div d<0 then=20
=
continue;<BR> =
=20
if dd[3] div d<0 then=20
=
continue;<BR> =
=20
writeln(dd[1] div d,' ',dd[2] div d,' ',dd[3] div d,'=20
=
',i);<BR> &nbs=
p;=20
=
close(output);<BR> &=
nbsp; =20
exit;<BR> =20
end;<BR> writeln('NONE');<BR> =20
close(output);<BR>end;<BR>begin<BR> =20
init;<BR> work;<BR>end.<BR></STRONG></P><STRONG>
<HR>
</STRONG>
<P></P>
<P></P>
=
<P><STRONG>=B5=DA=B6=FE=B8=F6=D3=C3=B5=BD=BE=D8=D5=F3=B3=CB=B7=A8</STRONG=
></P>
<P><STRONG>As there are only 101<SUP>3</SUP> =3D 1,030,301 ways to =
do this,=20
try them all and pick the best.</STRONG></P><PRE><STRONG>#include =
<stdio.h>
/* the goal ratio */
int goal[3];
/* the ratio of the feeds */
int ratios[3][3];
/* the best solution found so far */
int min;
int mloc[4]; /* amounts of ratio 1, 2, and 3, and the amount of ratio 4 =
prod */
/* the amount of each type of component in the feed */
int sum[3];
int main(int argc, char **argv)
{
FILE *fout, *fin;
int lv, lv2, lv3; /* loop variables */
int t, s; /* temporary variables */
int gsum; /* the sum of the amounts of components in the goal mixture =
*/
if ((fin =3D fopen("ratios.in", "r")) =3D=3D NULL)
{
perror ("fopen fin");
exit(1);
}
if ((fout =3D fopen("ratios.out", "w")) =3D=3D NULL)
{
perror ("fopen fout");
exit(1);
}
/* read in data */
fscanf (fin, "%d %d %d", &goal[0], &goal[1], &goal[2]);
for (lv =3D 0; lv < 3; lv++)
fscanf (fin, "%d %d %d", ratios[lv]+0, ratios[lv]+1, ratios[lv]+2);
gsum =3D goal[0] + goal[1] + goal[2];
min =3D 301; /* worst than possible =3D infinity */
/* boundary case (this ensures gsum !=3D 0) */
if (gsum =3D=3D 0)
{
fprintf (fout, "0 0 0 0\n");
return 0;
}
for (lv =3D 0; lv < 100; lv++)
for (lv2 =3D 0; lv2 < 100; lv2++)
{ /* for each amout of the first two types of mixtures */
sum[0] =3D lv*ratios[0][0] + lv2*ratios[1][0];
sum[1] =3D lv*ratios[0][1] + lv2*ratios[1][1];
sum[2] =3D lv*ratios[0][2] + lv2*ratios[1][2];
if (lv + lv2 > min) break;
for (lv3 =3D 0; lv3 < 100; lv3++)
{
s =3D lv + lv2 + lv3;
if (s >=3D min) break; /* worse than we've found already */
/* calculate a guess at the multiples of the goal we've obtained =
*/
/* use gsum so we don't have to worry about divide by zero */
t =3D (sum[0] + sum[1] + sum[2]) / gsum;
if (t !=3D 0 && sum[0] =3D=3D t*goal[0] &&=20
sum[1] =3D=3D t*goal[1] && sum[2] =3D=3D t*goal[2])
{ /* we've found a better solution! */
/* update min */
min =3D s;
/* store the solution */
mloc[0] =3D lv;
mloc[1] =3D lv2;
mloc[2] =3D lv3;
mloc[3] =3D t;
}
/* add another 'bucket' of feed #2 */
sum[0] +=3D ratios[2][0];
sum[1] +=3D ratios[2][1];
sum[2] +=3D ratios[2][2];
}
}
if (min =3D=3D 301) fprintf (fout, "NONE\n"); /* no solution found */
else fprintf (fout, "%d %d %d %d\n", mloc[0], mloc[1], mloc[2], =
mloc[3]);
return 0;
}</STRONG></PRE>
<H2>Vlad Novakovski's Solution</H2>
<P><STRONG>When you combine multiples of mixtures, you can look at =
it as a=20
multiplication of a matrix by a vector. For example, 8*(1:2:3) + =
1*(3:7:1)=20
+ 5*(2:1:2) =3D (21:28:35) =3D 7*(3:4:5) can be seen =
as</STRONG></P><PRE><STRONG>[ 1 3 2 ] [ 8 ] =20
[ 2 7 1 ] * [ 1 ] =3D 7 [3 4 5]
[ 3 1 2 ] [ 5 ]</STRONG></PRE>
<P><STRONG>The matrix and the goal ratio vector (3:4:5 in this =
case) are=20
given; what we have to find is the multiple vector (8:1:5 in this =
case)=20
and the proportionality costant (7 here). This is like solving a =
system of=20
linear equations. We can write it as</STRONG></P><PRE><STRONG>AX =
=3D kB.</STRONG></PRE>
<P><STRONG>Now, if we use Cramer's Rule, and let D =3D determinant =
of A,=20
then</STRONG></P><PRE><STRONG>X_1 =3D k D_1 / D
X_2 =3D k D_2 / D
X_3 =3D k D_3 / D,</STRONG></PRE>
<P><STRONG>where D_1 is the determinant of the matrix A with the =
1st=20
column is replaced by B, D_2 is the determinant of the matrix A =
with the=20
2nd column is replaced by B, D_3 is the determinant of the matrix =
A with=20
the 3rd column is replaced by B. (see a Linear Algebra textbook on =
why=20
this works.) ,P> We are looking for integral solutions. If D =
=3D 0, no=20
solutions. Otherwise, let k =3D D, and then X_1 =3D D_1, etc. If =
these values=20
(X_1,X_2,X_3, _and_ k) all have a greatest common factor above 1, =
divide=20
them all by that factor, as we are looking for the smallest =
possible=20
solutions.</STRONG></P>
<P><STRONG>Now, if some of the numbers is greater than 100, we =
have not=20
found a feasible solution, so we output `NONE'. Otherwise, the =
triple=20
(X_1,X_2,X_3) is output, as well as the value=20
k.</STRONG></P></DIV></TD></TR></TBODY></TABLE><BR>
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