1195: [HNOI2006]最短母串 - BZOJ

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

 

  1 var

  2     str:array[0..12,0..12]of string;

  3     g:array[0..12,0..12]of longint;

  4     f,fa:array[0..12,0..4096]of longint;

  5     a:array[0..12]of string;

  6     n:longint;

  7  

  8 function get(i,j:longint):string;

  9 var

 10     k,l:longint;

 11     flag:boolean;

 12 begin

 13     for k:=1 to length(a[i]) do

 14       begin

 15         if length(a[i])-k+1>length(a[j]) then continue;

 16         flag:=true;

 17         for l:=k to length(a[i]) do

 18           if a[i][l]<>a[j][l-k+1] then flag:=false;

 19         if flag then

 20         begin

 21           get:=a[i];

 22           for l:=length(a[i])-k+2 to length(a[j]) do

 23             get:=get+a[j][l];

 24           g[i,j]:=length(a[i])-k+1;

 25           exit(get);

 26         end;

 27       end;

 28     exit(a[i]+a[j]);

 29 end;

 30  

 31 procedure init;

 32 var

 33     i,j:longint;

 34 begin

 35     readln(n);

 36     for i:=1 to n do

 37       readln(a[i]);

 38     for i:=1 to n do

 39       for j:=1 to n do

 40         if i<>j then str[i,j]:=get(i,j);

 41 end;

 42  

 43 var

 44     d:array[0..49152,0..1]of longint;

 45  

 46 procedure work;

 47 var

 48     i,j,k,head,tail,min,lastj:longint;

 49 begin

 50     head:=1;

 51     tail:=0;

 52     for i:=1 to n do

 53       begin

 54         inc(tail);

 55         f[i,1<<(i-1)]:=length(a[i]);

 56         d[tail,0]:=i;

 57         d[tail,1]:=1<<(i-1);

 58       end;

 59     while head<=tail do

 60       begin

 61         for i:=1 to n do

 62           if d[head,1]and(1<<(i-1))=0 then

 63           begin

 64             if fa[i,d[head,1]+1<<(i-1)]=0 then

 65             begin

 66               inc(tail);

 67               d[tail,0]:=i;

 68               d[tail,1]:=d[head,1]+1<<(i-1);

 69               fa[i,d[head,1]+1<<(i-1)]:=d[head,0];

 70               f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];

 71             end;

 72             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

 73             begin

 74               f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];

 75               fa[i,d[head,1]+1<<(i-1)]:=d[head,0];

 76             end;

 77             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

 78             if str[i,fa[i,d[head,1]+1<<(i-1)]]>str[i,d[head,0]] then

 79             begin

 80               f[i,d[head,1]+1<<(i-1)]:=f[d[head,0],d[head,1]]+length(a[i])-g[i,d[head,0]];

 81               fa[i,d[head,1]+1<<(i-1)]:=d[head,0];

 82             end;

 83           end;

 84         inc(head);

 85       end;

 86     min:=1000;

 87     for i:=1 to n do

 88       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

 89       begin

 90         min:=f[i,1<<n-1];

 91         j:=i;

 92       end;

 93     write(a[j]);

 94     min:=1<<n-1;

 95     for i:=1 to n-1 do

 96       begin

 97         dec(min,1<<(j-1));

 98         lastj:=j;

 99         j:=fa[j,min+1<<(j-1)];

100         for k:=g[lastj,j]+1 to length(a[j]) do

101           write(a[j][k]);

102       end;

103 end;

104  

105 begin

106     init;

107     work;

108 end.

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