USACO Section 2.1 题解

 USACO 2.1.1 The Castle 城堡

此题便是Flood Fill了

 

{ID: lan jmPROG: castleLANG: PASCAL}program castle(input,output); const   q:array[1..4] of integer=(1,2,4,8);   qx:array[1..4] of integer=(0,-1,0,1);   qy:array[1..4] of integer=(-1,0,1,0); var   m,n,i,j,k,ans,max,a,b,c,x,y:longint;   f:array[0..51,0..51,1..4] of boolean;   map:array[0..51,0..51] of longint;   size:array[1..2500] of longint;procedure init; begin   assign(input,'castle.in');   assign(output,'castle.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;procedure bfs(xx,yy:longint);   //将相连的房间染上同种颜色 type   data=record     x,y:longint;    end; var   l,r,k,t:longint;   a:array[0..3000] of data; begin   l:=1;r:=1;a[1].x:=xx;a[1].y:=yy;t:=1;   while l<=r do     begin       for k:=1 to 4 do         if f[a[l].x,a[l].y,k] then           begin             xx:=a[l].x+qx[k];             yy:=a[l].y+qy[k];             if map[xx,yy]=0 then               begin                 map[xx,yy]:=ans;                 inc(t);                 inc(r);                 a[r].x:=xx;                 a[r].y:=yy;               end;           end;       inc(l);     end;   size[ans]:=t; end;begin  init;  readln(m,n);  fillchar(f,sizeof(f),true);  for i:=1 to n do    begin      for j:=1 to m do        begin          read(a);          for k:=4 downto 1 do            if a>=q[k] then              begin                f[i,j,k]:=false;                dec(a,q[k]);              end;            end;          readln;        end;  ans:=0;  for i:=1 to n do           //寻找未染色的房间开始染色    for j:=1 to m do      if map[i,j]=0 then        begin          inc(ans);          map[i,j]:=ans;          bfs(i,j);          if size[ans]>max then max:=size[ans];        end;  writeln(ans);  writeln(max);  for j:=1 to m do         //寻找移除哪面墙能得到的最大的房间    for i:=n downto 1 do      for k:=2 to 3 do        begin          x:=i+qx[k];          y:=j+qy[k];         if (map[x,y]<>0)and(map[i,j]<>map[x,y]) then           if max<size[map[x,y]]+size[map[i,j]] then             begin               max:=size[map[x,y]]+size[map[i,j]];               a:=i;               b:=j;               c:=k;             end;        end;  writeln(max);  write(a,' ',b,' ');  if c=2 then writeln('N')         else writeln('E');  outit;end.

 

 

USACO 2.1.2 Ordered Fractions 顺序的分数

此题用的是usaco的题解的解题方法，模拟下便知道了。其实这题暴力枚举也可以的，完全可以过的。

 

 

{ID: lan jmPROG: frac1LANG: PASCAL}program frac1(input,output); var   n:longint;procedure init; begin   assign(input,'frac1.in');   assign(output,'frac1.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;procedure work(a,b,c,d:longint); begin   if b+d>n then exit;   work(a,b,a+c,b+d);   writeln(a+c,'/',b+d);   work(a+c,b+d,c,d); end;begin  init;  readln(n);  writeln('0/1');  work(0,1,1,1);  writeln('1/1');  outit;end.

 

 

 

USACO 2.1.3 Sorting a Three-Valued Sequence 三值的排序

模拟....

 

 

{ID: lan jmPROG: sort3LANG: PASCAL}program sort3(input,output); var   n,l1,l2,i,ans,x,y:longint;   a:array[1..10000] of 1..3;procedure init; begin   assign(input,'sort3.in');   assign(output,'sort3.out');   reset(input);   rewrite(output); end;procedure outit;  begin   close(input);   close(output); end;begin  init;  readln(n);  for i:=1 to n do    begin      readln(a[i]);      if a[i]=1 then inc(l1);      if a[i]=2 then inc(l2);    end;  for i:=1 to l1 do                 //数值为1的段    begin      if a[i]<>1 then inc(ans);      if a[i]=3 then inc(x);    end;    for i:=l1+1 to l1+l2 do            //数值为2的段    if a[i]=3 then inc(ans);    for i:=l1+l2+1 to n do            //数值为3的段    if a[i]=1 then inc(y);    if x>y then writeln(ans+x-y)         else writeln(ans);  outit;end.

 

 

 

USACO 2.1.4 Healthy Holsteins健康的好斯坦奶牛

 

此题我刚开始看以为是01背包，可是维数过多行不通，然后看数据极小，就dfs了...

 

 

{ID: lan jmPROG: holsteinLANG: PASCAL}program holstein(input,output);var  i,j,v,g,ans:longint;  answer,f:array[1..15] of boolean;  a:array[1..25] of longint;  b:array[1..15,1..25] of longint;procedure init; begin   assign(input,'holstein.in');   assign(output,'holstein.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;procedure work;var  i,j,k:longint;  p:array[1..25] of longint;begin  k:=0;  for i:=1 to g do    if f[i] then inc(k);  if k>=ans then exit;  fillchar(p,sizeof(p),0);  for i:=1 to  g do    if f[i] then      for j:=1 to v do        inc(p[j],b[i,j]);  for i:=1 to v do    if p[i]<a[i] then exit;  ans:=k;  for i:=1 to g do answer[i]:=f[i];end;procedure dfs(dep:longint); begin   if dep=g+1 then     begin       work;       exit;     end;   f[dep]:=true;   dfs(dep+1);   f[dep]:=false;   dfs(dep+1); end;begin  init;  readln(v);  for i:=1 to v do read(a[i]);  readln(g);  for i:=1 to g do    begin      for j:=1 to v do read(b[i,j]);      readln;    end;  ans:=maxlongint;  dfs(1);  write(ans);  for i:=1 to g do    if answer[i] then write(' ',i);  writeln;  outit;end.

 

USACO 2.1.5 Hamming Codes海明码

 

看清题目很重要！（两两、至少）

 

{ID: lan jmPROG: hammingLANG: PASCAL}program hamming(input,output);const  wei:array[1..8] of integer=(1,3,7,15,31,63,127,255);  //二进制位数var  n,b,d,i,ans:longint;  a:array[1..64] of longint;procedure init; begin   assign(input,'hamming.in');   assign(output,'hamming.out');   reset(input);   rewrite(output); end;procedure outit; begin   close(input);   close(output); end;function ok(x:longint):boolean;var  t,i,s:longint;begin  for i:=1 to ans do    begin      t:=0;      s:=x xor a[i];      while s<>0 do        begin          if s mod 2=1 then inc(t);          s:=s div 2;        end;      if t<d then exit(false);    end;  exit(true);end;begin  init;  readln(n,b,d);  a[1]:=0;  ans:=1;  for i:=1 to wei[b] do    if ans<n then      if ok(i) then        begin          inc(ans);          a[ans]:=i;        end;  for i:=1 to ans-1 do      if i mod 10<>0 then write(a[i],' ')                     else writeln(a[i]);  writeln(a[ans]);  outit;end.