【字典序+栈】接龙游戏(words.pas/in/out)

 

接龙游戏

(words.pas/in/out)

Problem

给出n个单词,已经按长度排好了序.如果单词I是单词J的前缀,那么I- J算一次接龙.

你的任务是:对于输入的单词,找出最长的龙.

Input

数据第一行为n(1<=n<=100000),以下n行每行是一个单词(由小写组成),已经按长度排好了序.每个单词的长度不超过50.

Output

输出一行，这行为一个数,即最长的龙的长度.

Sample Input

5

i

a

int

able

inter

Sample Input

3

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这道题一测的时候，我全超时

那时我是直接用DP..并没有任何的排序..

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其实这道题可以先按字典序排了之后再维护一个栈就可以了..

以后要记住了...

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考试时还想到了一种方法..==做..

建一棵树...

保证左子树并不是根点的后缀，右子树为根节点的后缀...

 

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var  n:longint;  s,s_z:array[1..100000]of string;  s_t:longint;  ans:longint; // f:array[1..100000]of longint;procedure init;begin  assign(input,'words.in');  assign(output,'words.out');  reset(input); rewrite(output);end;procedure terminate;begin  close(input); close(output);  halt;end;function pd(x,y:longint):boolean;var  i:longint;begin  pd:=true;  for i:=1 to length(s[x]) do    if s[x][i]<>s[y][i] then exit(false);end;procedure qsort(l,r:longint);var  i,j:longint;  x,tem:string;begin  i:=l; j:=r;  x:=s[(l+r)shr 1];  repeat    while x<s[j] do dec(j);    while s[i]<x do inc(i);    if i<=j then      begin        tem:=s[i];        s[i]:=s[j];        s[j]:=tem;        inc(i); dec(j);      end;  until i>j;  if i<r then qsort(i,r);  if l<j then qsort(l,j);end;function pd(s1,s2:string):boolean;var  i:longint;begin  pd:=true;  for i:=1 to length(s1) do    if s1[i]<>s2[i] then exit(false);end;procedure ru(i:longint);begin  if s_t=0 then    begin      inc(s_t);      s_z[s_t]:=s[i];      if s_t>ans then ans:=s_t;    end    else    begin      while (s_t>=1) and not pd(s_z[s_t],s[i]) do dec(s_t);      inc(s_t);      s_z[s_t]:=s[i];      if s_t>ans then ans:=s_t;    end;end;procedure main;var  i:longint;begin  readln(n);  for i:=1 to n do    begin      readln(s[i]);     // f[i]:=1;    end;  qsort(1,n);  s_t:=0;  ans:=0;  for i:=1 to n do    begin      ru(i);    end;  writeln(ans);end;begin  init;  main;  terminate;end.