[POJ1159 Palindrome]

[题目来源]：IOI 2000

[关键字]：动态规划

[题目大意]：把一个字符串变成回文串所需的最小步骤。

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

[分析]：因为回文串的要求是正着读和倒着读是一样的，即把这个字符串倒置后进行匹配，用动态规划求出所需最小操作数。f[i,j]={f[i-1,j-1]|s1[i]=s2[j],min{f[i-1,j],f[i,j-1]+1|s1[i]<>s2[j]}}。但是最后要除以二，因为两个字符串其实是同一个，所以所有操作都会被计算两次（自己模拟一下）。

[代码]：

View Code

 1 program Project1;
 2 var
 3   n: longint;
 4   f: array[0..5010,0..5010] of integer;
 5   s1, s2: ansistring;
 6 
 7 procedure init;
 8 var
 9   i: longint;
10 begin
11   readln(n);
12   readln(s1);
13   s2 := '';
14   for i := n downto 1 do
15     s2 := s2+s1[i];
16   //writeln(s1,'#',s2,'@');
17   //readln;
18   //readln;
19 end;
20 
21 function min(x, y: longint):longint;
22 begin
23   if x < y then exit(x) else exit(y);
24 end;
25 
26 procedure work;
27 var
28   i, j, st: longint;
29 begin
30   fillchar(f,sizeof(f),0);
31   for i := 1 to n do
32     begin
33       f[i,0] := i;
34       f[0,i] := i;
35     end;
36   for i := 1 to n do
37     for j := 1 to n do
38       begin
39         if s1[i] = s2[j] then begin f[i,j] := f[i-1,j-1]; continue; end;
40         f[i,j] := min(f[i-1,j],f[i,j-1])+1;
41       end;
42   writeln(f[n,n] shr 1);
43   readln;
44 end;
45 
46 begin
47   init;
48   work;
49 end.