1195: [HNOI2006]最短母串

Description

给定n个字符串（S1,S2,„,Sn），要求找到一个最短的字符串T，使得这n个字符串（S1,S2,„,Sn）都是T的子串。
Input

第一行是一个正整数n（n<=12），表示给定的字符串的个数。以下的n行，每行有一个全由大写字母组成的字符串。每个字符串的长度不超过50.
Output

只有一行，为找到的最短的字符串T。在保证最短的前提下，如果有多个字符串都满足要求，那么必须输出按字典序排列的第一个。
Sample Input

2

ABCD

BCDABC
Sample Output

ABCDABC

状压DP，f[i,s]表示以i开头状态为s的情况最短字符串有多长

为了比较字典序和方便的输出最终串，我们还要记录fa[i,s]表示f[i,s]的第二个字符串是哪个

var
    str:array[0..12,0..12]of string;
    g:array[0..12,0..12]of longint;
    f,fa:array[0..12,0..4096]of longint;
    a:array[0..12]of string;
    n:longint;
 
function get(i,j:longint):string;
var
    k,l:longint;
    flag:boolean;
begin
    for k:=1 to length(a[i]) do
      begin
        if length(a[i])-k+1>length(a[j]) then continue;
        flag:=true;
        for l:=k to length(a[i]) do
          if a[i][l]<>a[j][l-k+1] then flag:=false;
        if flag then
        begin
          get:=a[i];
          for l:=length(a[i])-k+2 to length(a[j]) do
            get:=get+a[j][l];
          g[i,j]:=length(a[i])-k+1;
          exit(get);
        end;
      end;
    exit(a[i]+a[j]);
end;
 
procedure init;
var
    i,j:longint;
begin
    readln(n);
    for i:=1 to n do
      readln(a[i]);
    for i:=1 to n do
      for j:=1 to n do
        if i<>j then str[i,j]:=get(i,j);
end;
 
var
    d:array[0..49152,0..1]of longint;
 
procedure work;
var
    i,j,k,head,tail,min,lastj:longint;
begin
    head:=1;
    tail:=0;
    for i:=1 to n do
      begin
        inc(tail);
        f[i,1<<(i-1)]:=length(a[i]);
        d[tail,0]:=i;
        d[tail,1]:=1<<(i-1);
      end;
    while head<=tail do
      begin
        for i:=1 to n do
          if d[head,1]and(1<<(i-1))=0 then
          begin
            if fa[i,d[head,1]+1<<(i-1)]=0 then
            begin
              inc(tail);
              d[tail,0]:=i;
              d[tail,1]:=d[head,1]+1<<(i-1);
              fa[i,d[head,1]+1<<(i-1)]:=d[head,0];
              f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];
            end;
            if f[i,d[head,1]+1<<(i-1)]>f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]] then
            begin
              f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];
              fa[i,d[head,1]+1<<(i-1)]:=d[head,0];
            end;
            if f[i,d[head,1]+1<<(i-1)]=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]] then
            if str[i,fa[i,d[head,1]+1<<(i-1)]]>str[i,d[head,0]] then
            begin
              f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];
              fa[i,d[head,1]+1<<(i-1)]:=d[head,0];
            end;
          end;
        inc(head);
      end;
    min:=1000;
    for i:=1 to n do
      if (f[i,1<<n-1]<min)or((f[i,1<<n-1]=min)and(str[i,fa[i,1<<n-1]]<str[j,fa[j,1<<n-1]])) then
      begin
        min:=f[i,1<<n-1];
        j:=i;
      end;
    write(a[j]);
    min:=1<<n-1;
    for i:=1 to n-1 do
      begin
        dec(min,1<<(j-1));
        lastj:=j;
        j:=fa[j,min+1<<(j-1)];
        for k:=g[lastj,j]+1 to length(a[j]) do
          write(a[j][k]);
      end;
end;
 
begin
    init;
    work;
end.