usaco 1.4 clocks 2008.11.5

usaco 1.4 clocks 2008.11.5

小结

1.clocks-宽搜.pas  宽搜，只过了两组，就溢出了，本来想开大一点队列，但占用的空间太多，超出了测试的要求，又看到网上的人也用的宽搜，也是只过了两组，就放弃了宽搜。

2.clocks-meiju.pas  然后看到了网上的结题报告，如下：由于操作4次等于没操作，所以只用枚举操作0次到操作3次这几种情况，枚举此时在最差情况下也只有4^9次，可以接受，然后就编了个9重循环，ac了。只是要的打字比较多，数组也不能做循环变量

program1：宽搜2组

{ID:  PROG: clocksLANG: PASCAL}program p_clocks;const  maxx=10000;  maxn=300;  fin='clocks.in';fout='clocks.out';  a:array[1..9,0..5]of integer=         ((4,1,2,4,5,0),(3,1,2,3,0,0),(4,2,3,5,6,0),          (3,1,4,7,0,0),(5,2,4,5,6,8),(3,3,6,9,0,0),          (4,4,5,7,8,0),(3,7,8,9,0,0),(4,5,6,8,9,0));type  arr=array[1..9]of integer;  arr2=array[0..maxn]of integer;  code=record       m:arr;       v:arr2;       end;var  y:array[1..maxx]of code;  p:arr;  r,g:arr2;  f1,f2:text;  i,x:integer;  f:array[0..3,0..3,0..3,0..3,0..3,0..3,0..3,0..3,0..3]of boolean;     procedure init;     var i,k:integer;     begin        assign(f1,fin);reset(f1);        assign(f2,fout);rewrite(f2);         for i:=1 to 9 do               begin read(f1,p[i]);k:=(p[i] div 3)mod 4;p[i]:=k; end;     end;     procedure deal;     var i,u,j:integer;c:arr;s:arr2;     begin     u:=0;     repeat inc(u);         for i:=1 to 9 do            begin                c:=y[u].m;s:=y[u].v;                for j:=1 to a[i,0] do                     c[a[i,j]]:=(c[a[i,j]]+1)mod 4;                inc(s[0]);s[s[0]]:=i;                if c[1]+c[2]+c[3]+c[4]+c[5]+c[6]+c[7]+c[8]+c[9]=0 then                    begin g:=s;exit;end else                       if f[c[1],c[2],c[3],c[4],c[5],c[6],c[7],c[8],c[9]] then                          begin inc(x);y[x mod maxx].m:=c;y[x mod maxx].v:=s;                              f[c[1],c[2],c[3],c[4],c[5],c[6],c[7],c[8],c[9]]:=false;                            end;           end;      until u>=x;      end;     {-------------main--------------}     begin         fillchar(f,sizeof(f),true);       init;       fillchar(r,sizeof(r),0);       x:=1;y[1].m:=p;y[1].v:=r;       deal;       for i:=1 to g[0]-1 do         write(f2,g[i],' ');         writeln(f2,g[g[0]]);        close(f1);        close(f2);      end.

program2：九重循环：有时枚举也是不错的方法。

{ID: PROG: clocksLANG: PASCAL}{}program p_clock_meiju;  const  fin='clocks.in';fout='clocks.out';  a:array[1..9,0..5]of integer=         ((4,1,2,4,5,0),(3,1,2,3,0,0),(4,2,3,5,6,0),          (3,1,4,7,0,0),(5,2,4,5,6,8),(3,3,6,9,0,0),          (4,4,5,7,8,0),(3,7,8,9,0,0),(4,5,6,8,9,0));  type     arr=array[1..9]of integer;  var  f1,f2:text;  p,e:arr;  g:array[1..30]of longint;  g1,i,j,t1,t2,t3,t4,t5,t6,t7,t8,t9:integer;  procedure init;     var i,k:integer;     begin        assign(f1,fin);reset(f1);        assign(f2,fout);rewrite(f2);        for i:=1 to 9 do           begin read(f1,p[i]);k:=(p[i] div 3)mod 4;p[i]:=k;end;     end;procedure print;begin  g1:=0;  while t1<>0 do     begin dec(t1);inc(g1);g[g1]:=1;end;    while t2<>0 do     begin dec(t2);inc(g1);g[g1]:=2;end;    while t3<>0 do     begin dec(t3);inc(g1);g[g1]:=3;end;     while t4<>0 do     begin dec(t4);inc(g1);g[g1]:=4;end;      while t5<>0 do     begin dec(t5);inc(g1);g[g1]:=5;end;       while t6<>0 do     begin dec(t6);inc(g1);g[g1]:=6;end;       while t7<>0 do     begin dec(t7);inc(g1);g[g1]:=7;end;       while t8<>0 do     begin dec(t8);inc(g1);g[g1]:=8;end;       while t9<>0 do     begin dec(t9);inc(g1);g[g1]:=9;end;     for i:=1 to g1-1 do write(f2,g[i],' ');     writeln(f2,g[g1]);     end;procedure doit;var u,k,w:integer;  begin     for t1:=0 to 3 do       for t2:=0 to 3 do         for t3:=0 to 3 do           for t4:=0 to 3 do             for t5:=0 to 3 do               for t6:=0 to 3 do                 for t7:=0 to 3 do                   for t8:=0 to 3 do                     for t9:=0 to 3 do                         begin e:=p;                            for k:=1 to a[1,0] do                                  e[a[1,k]]:=e[a[1,k]]+t1;                             for k:=1 to a[2,0] do                                       e[a[2,k]]:=e[a[2,k]]+t2;                             for k:=1 to a[3,0] do                                     e[a[3,k]]:=e[a[3,k]]+t3;                               for k:=1 to a[4,0] do                                     e[a[4,k]]:=e[a[4,k]]+t4;                                for k:=1 to a[5,0] do                                    e[a[5,k]]:=e[a[5,k]]+t5;                                 for k:=1 to a[6,0] do                                    e[a[6,k]]:=e[a[6,k]]+t6;                                 for k:=1 to a[7,0] do                                    e[a[7,k]]:=e[a[7,k]]+t7;                                  for k:=1 to a[8,0] do                                     e[a[8,k]]:=e[a[8,k]]+t8;                                  for k:=1 to a[9,0] do                                    e[a[9,k]]:=e[a[9,k]]+t9;                                   u:=0;                            for j:=1 to 9 do                                 begin                                    w:=e[j] mod 4;                                    e[j]:=w;                                    inc(u,e[j]);                                 end;                             if u=0 then begin print;exit;end;                         end;     end;     {--------------main----------------}     begin       init;       doit;       close(f1);       close(f2);     end.

data

------- test 1 -------

9 9 12

6 6 6

6 3 6

------- test 2 -------

12 9 12

9 9 9

12 9 12

------- test 3 -------

6 9 3

3 3 9

12 12 12

------- test 4 -------

9 3 9

9 9 9

9 9 9

------- test 5 -------

6 12 12

12 12 12

12 12 12

------- test 6 -------

3 12 9

6 6 6

12 12 12

------- test 7 -------

12 3 3

3 6 6

12 3 6

------- test 8 -------

12 3 9

9 12 12

3 6 9

------- test 9 -------

9 12 9

12 3 12

9 12 9