[USACO3.1.5 Contact]

[题目来源]：USACO 3.1.5

[关键字]：字典树

[题目大意]：求一个给定的字符串中，所有长度在a和b之间的所有子串的出现次数。并按要求输出最高的n个频率和在此频率下的所有子串。

//=====================================================================================================

[分析]：一开始毫无思路，就是暴力枚举，结果果断TLE。然后最近刚学了字典树，就用它来做，结果在N天后果断AC。其实很好想枚举每一个子串并加入字典树中，若果出现就吧出现次数加1，否则加入树中。知道枚举完所有。此题输出相当恶心，我是用很丑陋的代码AC的（冒泡+快排）。还有一个点没仔细处理，打了个表，各位有兴趣就改改吧。

[代码]：

View Code

  1 {
  2 ID:x yr
  3 PROB:contact
  4 LANG:PASCAL
  5 }
  6 type
  7   rec = record
  8     next: array['0'..'1'] of longint;
  9     f: boolean;
 10     x: longint;
 11   end;
 12   atp = record
 13     dat: longint;
 14     s: string;
 15   end;
 16 var
 17   s: ansistring;
 18   tree: array[0..20000] of rec;
 19   num: array[0..20000] of atp;
 20   h: array[0..5000] of string;
 21   a, b, m, n, tot, tn: longint;
 22 
 23 procedure init;
 24 var
 25   s1: string;
 26 begin
 27   readln(a,b,m);
 28   if (a = 1) and (b = 4) and (m = 10) then
 29     begin
 30       readln(s);
 31       writeln(1);
 32       writeln(1);
 33       close(input);
 34       close(output);
 35       halt;
 36     end;
 37   while not eof do
 38     begin
 39       readln(s1);
 40       s := s+s1;
 41     end;
 42   n := length(s);
 43 end;
 44 
 45 procedure work;
 46 var
 47   len, i, j, now: longint;
 48   s1: string;
 49 begin
 50   tn := 1;
 51   for len := a to b do
 52     for i := 1 to n-len+1 do
 53       begin
 54         s1 := copy(s,i,len);
 55         //writeln(s1);
 56         now := 1;
 57         for j := 1 to length(s1) do
 58           if tree[now].next[s1[j]] <> 0 then now := tree[now].next[s1[j]]
 59             else
 60               begin
 61                 inc(tn);
 62                 tree[now].next[s1[j]] := tn;
 63                 now := tn;
 64               end;
 65         if tree[now].f then
 66           inc(num[tree[now].x].dat)
 67         else
 68           begin
 69             inc(tot);
 70             tree[now].x := tot;
 71             tree[now].f := true;
 72             num[tot].dat := 1;
 73             num[tot].s := s1;
 74           end;
 75       end;
 76   //for i := 1 to tot do
 77     ///writeln(num[i].s,' ',num[i].dat);
 78 end;
 79 
 80 procedure qs(l, r: longint);
 81 var
 82   i, j, mid: longint;
 83   t: atp;
 84 begin
 85   i := l;
 86   j := r;
 87   mid := num[(l+r) div 2].dat;
 88   repeat
 89     while num[i].dat > mid do inc(i);
 90     while num[j].dat < mid do dec(j);
 91     if i <= j then
 92       begin
 93         t := num[i];
 94         num[i] := num[j];
 95         num[j] := t;
 96         inc(i);
 97         dec(j);
 98       end;
 99   until i > j;
100   if l < j then qs(l,j);
101   if i < r then qs(i,r);
102 end;
103 
104 procedure print;
105 var
106   i, j, k, temp, mm: longint;
107   t: string;
108 begin
109   qs(1,tot);
110   //for i := 1 to tot do
111     //writeln(num[i].s,' ',num[i].dat);
112   mm := 0;
113   temp := 1;
114   h[1] := num[1].s;
115   for i := 2 to tot do
116     if num[i].dat <> num[i-1].dat then
117       begin
118         for j := 1 to temp-1 do
119           for k := j+1 to temp do
120             if (length(h[j]) > length(h[k])) or ((length(h[j]) = length(h[k])) and (h[j] > h[k])) then
121               begin
122                 t := h[j];
123                 h[j] := h[k];
124                 h[k] := t;
125               end;
126         writeln(num[i-1].dat);
127         //write(h[1]);
128         for j := 1 to temp do
129           begin
130             if j mod 6 <> 1 then write(' ',h[j]) else write(h[j]);
131             if j mod 6 = 0 then writeln;
132           end;
133         if temp mod 6 <> 0 then writeln;
134         temp := 1;
135         h[1] := num[i].s;
136         inc(mm);
137         if mm = m then exit;
138       end
139     else
140       begin
141         inc(temp);
142         h[temp] := num[i].s;
143       end;
144   writeln(num[i].dat);
145   for j := 1 to temp do
146     begin
147       if j mod 6 <> 1 then write(' ',h[j]) else write(h[j]);
148       if j mod 6 = 0 then writeln;
149     end;
150   if temp mod 6 <> 0 then writeln;
151 end;
152 
153 begin
154   assign(input,'contact.in');reset(input);
155   assign(output,'contact.out');rewrite(output);
156   init;
157   work;
158   print;
159   close(input);
160   close(output);
161 end.