[NDK 双栈排序]

[题目来源]：NOIP2004

[关键字]：二分图

[题目大意]：判断是否能用两个栈将给定的序列排列成升序，如果能则输出操作序列。

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[分析]：易证对于i<j如果存在k，使i<j<k且a[k]<a[i]<a[j]则i和j必定不能在同一个栈里，所以将他们连一条无向边。然后就是利用染色法判断是否是一个二分图。染色时吧标号小的放进栈1就能得出字典序小的操作序列。

[代码]：

View Code

  1 var
  2   a, b: array[0..2000] of longint;
  3   s: array[0..2000] of -1..2;
  4   s1, s2: array[0..2000] of longint;
  5   n, t1, t2: longint;
  6 
  7 procedure init;
  8 var
  9   i: longint;
 10 begin
 11   readln(n);
 12   for i := 1 to n do read(a[i]);
 13   for i := n downto 1 do
 14     begin
 15       b[i] := a[i];
 16       if (i < n) and (b[i] > b[i+1]) then b[i] := b[i+1];
 17       s[i] := -1;
 18     end;
 19   //for i := 1 to n do write(b[i],'');
 20   //writeln;
 21 end;
 22 
 23 procedure make(x, y: longint);
 24 begin
 25   if (s[x] = -1) and (s[y] <> -1) then
 26     begin s[x] := 3-s[y]; exit; end;
 27   if (s[x] <> -1) and (s[y] = -1) then
 28     begin s[y] := 3-s[x]; exit; end;
 29   if (s[x] = -1) and (s[y] = -1) then
 30     begin s[x] := 1; s[y] := 2; exit; end;
 31   if (s[x] <> -1) and (s[y] <> -1) then
 32     if s[x] = s[y] then
 33       begin
 34         writeln(0);
 35         halt;
 36       end;
 37 end;
 38 
 39 procedure print;
 40 var
 41   i, r: longint;
 42 begin
 43   r := 1;
 44   t1 := 0;
 45   t2 := 0;
 46   for i := 1 to n do
 47     begin
 48       while (s1[t1] = r) or (s2[t2] = r) do
 49         begin
 50           if s1[t1] = r then
 51             begin
 52               write('b ');
 53               dec(t1);
 54               inc(r);
 55             end;
 56           if s2[t2] = r then
 57             begin
 58               write('d ');
 59               dec(t2);
 60               inc(r);
 61             end;
 62         end;
 63       if s[i] = 1 then
 64         begin
 65           write('a ');
 66           inc(t1);
 67           s1[t1] := a[i];
 68         end;
 69       if s[i] = 2 then
 70         begin
 71           write('c ');
 72           inc(t2);
 73           s2[t2] := a[i];
 74         end;
 75     end;
 76   while (s1[t1] = r) or (s2[t2] = r) do
 77     begin
 78       if s1[t1] = r then
 79         begin
 80           write('b ');
 81           dec(t1);
 82           inc(r);
 83         end;
 84       if s2[t2] = r then
 85         begin
 86           write('d ');
 87           dec(t2);
 88           inc(r);
 89         end;
 90     end;
 91 end;
 92 
 93 procedure work;
 94 var
 95   i, j: longint;
 96 begin
 97   for i := 1 to n-2 do
 98     for j := i+1 to n-1 do
 99       if (a[i] < a[j]) and (b[j+1] < a[i]) then make(i,j);
100   for i := 1 to n do
101     if s[i] = -1 then s[i] := 1;
102 end;
103 
104 begin
105   assign(input,'d:\1.in');reset(input);
106   assign(output,'d:\1.out');rewrite(output);
107   init;
108   work;
109   print;
110   close(input);
111   close(output);
112 end.