usaco 2_2_4 party lamps 题解_leokan的blog.mht

美国USACO题库源程序

bsp; =20
      =
break;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      if not haven=20
      =
then<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
inc(t);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
queue[t]:=3Dtemp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
count[t]:=3Dcount[h]+1;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      =
temp:=3Dqueue[h];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;=20
      temp:=3Dtemp xor (1 shl 6=20
      =
-1);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      =
haven:=3Dfalse;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=
&nbsp;&nbsp;=20
      for i:=3D1 to t=20
      =
do<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      if temp=3Dqueue[i]=20
      =
then<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      =
haven:=3Dtrue;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;=20
      =
break;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      if not haven=20
      =
then<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
inc(t);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
queue[t]:=3Dtemp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
count[t]:=3Dcount[h]+1;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      inc(h);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      end;<BR>end;</STRONG></P>
      <P></P>
      <P><BR><STRONG>procedure work;<BR>var<BR>&nbsp;&nbsp;&nbsp;=20
      i,j,temp:integer;<BR>&nbsp;&nbsp;&nbsp;=20
      noon,found:boolean;<BR>&nbsp;&nbsp;&nbsp;=20
      s,st:string;<BR>begin<BR>&nbsp;&nbsp;&nbsp;=20
      assign(output,'lamps.out');rewrite(output);<BR>&nbsp;&nbsp;&nbsp;=20
      found:=3Dfalse;<BR>&nbsp;&nbsp;&nbsp; bfs;<BR>&nbsp;&nbsp;&nbsp;=20
      noon:=3Dtrue;<BR>&nbsp;&nbsp;&nbsp; while noon=20
      =
do{=B6=D4=CB=F9=D3=D0=C7=E9=BF=F6=C3=B0=C5=DD=C5=C5=D0=F2}<BR>&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;=20
      =
noon:=3Dfalse;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;=20
      for i:=3D1 to t-1=20
      =
do<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      if queue[i]&gt;queue[i+1]=20
      =
then<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      =
temp:=3Dqueue[i];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;=20
      =
queue[i]:=3Dqueue[i+1];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;=20
      =
queue[i+1]:=3Dtemp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;=20
      =
temp:=3Dcount[i];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;=20
      =
count[i]:=3Dcount[i+1];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;&nbsp;=20
      =
count[i+1]:=3Dtemp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;=20
      =
noon:=3Dtrue;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      end;<BR>&nbsp;&nbsp;&nbsp; for i:=3D1 to t=20
      =
do{=C3=B6=BE=D9=CB=F9=D3=D0=C7=E9=BF=F6}<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;=20
      if (count[i]&gt;c) then=20
      =
continue;{=B3=AC=B9=FD=BF=AA=B9=D8=B4=CE=CA=FD}<BR>&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      if (queue[i] and check1&lt;&gt;check1)or(queue[i] or =
check2&lt;&gt;check2)=20
      then=20
      =
continue;{=B2=BB=B7=FB=BA=CF=CF=DE=D6=C6}<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      =
found:=3Dtrue;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;=20
      =
s:=3D'';<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;=20
      for j:=3D1 to 6=20
      =
do<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      =
begin<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      if queue[i] and 1&gt;0 then s:=3Ds+'1' else=20
      =
s:=3Ds+'0';{=D7=AA=CE=AA2=BD=F8=D6=C6}<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;=20
      queue[i]:=3Dqueue[i] shr=20
      =
1;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      =
end;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;=20
      =
st:=3D'';<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=
&nbsp;=20
      for j:=3D6 downto 1=20
      =
do<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      =
st:=3Dst+s[j];<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;=20
      for j:=3D1 to n=20
      =
do<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp;&nbsp;&nbsp;=20
      write(st[index[j mod=20
      =
6]]);<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;=20
      writeln;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
      end;<BR>&nbsp;&nbsp;&nbsp; if not found then=20
      writeln('IMPOSSIBLE');<BR>&nbsp;&nbsp;&nbsp;=20
      close(output);<BR>end;<BR>begin<BR>&nbsp;&nbsp;&nbsp;=20
      init;<BR>&nbsp;&nbsp;&nbsp; work;<BR>end.</STRONG></P><STRONG>
      <HR>
      </STRONG>
      <P><STRONG>USACO=B5=C4=B7=D6=CE=F6</STRONG></P>
      <P><STRONG>There are a number of insights required for this =
problem.=20
      </STRONG></P>
      <P><STRONG>The main insight is that no matter which switches get =
pressed,=20
      the light pattern will repeat every six lamps. Another insight is =
that the=20
      order of switch presses does not matter, and that pressing a =
switch twice=20
      is the same as not pressing a switch at all. </STRONG></P>
      <P><STRONG>Any even number of switch presses greater than four =
might as=20
      well just be four switch presses, and and any odd number greater =
than=20
      three might as well just be three presses. </STRONG></P>
      <P><STRONG>We can treat the light information as describing only =
the first=20
      six lamps (by treating lamp n as lamp (n mod 6)), and only try =
pressing=20
      each switch once or not at all. </STRONG></P>
      <P><STRONG>To maintain the actual information about which lights =
are on,=20
      we use an integer holding six bits. The least significant bit is =
light 6,=20
      the next least is light 5, and so on. This has the effect that =
treating=20
      these bit strings as normal numbers sorts the same way that we =
need to=20
      print them. </STRONG></P>
      <P><STRONG>Switches are represented as bit vectors that say which =
lights=20
      get toggled. </STRONG></P><PRE><STRONG>#include &lt;stdio.h&gt;
#include &lt;stdlib.h&gt;
#include &lt;string.h&gt;
#include &lt;assert.h&gt;

#define MAXLAMP 6
#define LAMPMASK ((1&lt;&lt;MAXLAMP)-1)

int nlamp;
int nswitch;
int ison;
int known;

int poss[1&lt;&lt;MAXLAMP];

int flip[4] =3D {
    LAMPMASK,  /* flip all lights */
    LAMPMASK &amp; 0xAA,  /* flip odd lights */
    LAMPMASK &amp; 0x55, /* flip even lights */
    LAMPMASK &amp; ((1&lt;&lt;(MAXLAMP-1))|(1&lt;&lt;(MAXLAMP-4))) /* =
lights 1, 4 */
};

/*
 * Starting with current light state ``lights'', flip exactly n switches
 * with number &gt;=3D i.
 */
void
search(int lights, int i, int n)
{
    if(n =3D=3D 0) {
 if((lights &amp; known) =3D=3D ison)
     poss[lights] =3D 1;
 return;
    }

    for(; i&lt;4; i++)
 search(lights ^ flip[i], i+1, n-1);
}

void
printseq(FILE *fout, int lights)
{
    int i;
    char s[100+1];

    for(i=3D0; i&lt;nlamp; i++)
 s[i] =3D (lights &amp; (1&lt;&lt;(MAXLAMP-1 - i%MAXLAMP))) ? '1' : '0';
    s[nlamp] =3D '\0';
    fprintf(fout, "%s\n", s);
}

void
main(void)
{
    FILE *fin, *fout;
    int a, i, impossible;

    fin =3D fopen("lamps.in", "r");
    fout =3D fopen("lamps.out", "w");
    assert(fin !=3D NULL &amp;&amp; fout !=3D NULL);

    fscanf(fin, "%d %d", &amp;nlamp, &amp;nswitch);

    for(;;) {
 fscanf(fin, "%d", &amp;a);
 if(a =3D=3D -1)
     break;
 a =3D MAXLAMP-1 - (a-1) % MAXLAMP;
 ison |=3D 1&lt;&lt;a;
 known |=3D 1&lt;&lt;a;
    }

    for(;;) {
 fscanf(fin, "%d", &amp;a);
 if(a =3D=3D -1)
     break;
 a =3D MAXLAMP-1 - (a-1) % MAXLAMP;
 assert((ison &amp; (1&lt;&lt;a)) =3D=3D 0);
 known |=3D 1&lt;&lt;a;
    }

    if(nswitch &gt; 4)
 if(nswitch%2 =3D=3D 0)
     nswitch =3D 4;
 else
     nswitch =3D 3;

    for(; nswitch &gt;=3D 0; nswitch -=3D 2)
     search(LAMPMASK, 0, nswitch);

    impossible =3D 1;
    for(i=3D0; i&lt;(1&lt;&lt;MAXLAMP); i++) {
 if(poss[i]) {
     printseq(fout, i);
     impossible =3D 0;
 }
    }
    if(impossible)
 fprintf(fout, "IMPOSSIBLE\n");

    exit(0);
}</STRONG></PRE></DIV></TD></TR></TBODY></TABLE><BR>
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