[动态规划]Pku1159--Palindrome

http://acm.pku.edu.cn/JudgeOnline/problem?id=1159

 

Palindrome

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 27432 Accepted: 9174

Description

A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome.

As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.

Input

Your program is to read from standard input. The first line contains one integer: the length of the input string N, 3 <= N <= 5000. The second line contains one string with length N. The string is formed from uppercase letters from 'A' to 'Z', lowercase letters from 'a' to 'z' and digits from '0' to '9'. Uppercase and lowercase letters are to be considered distinct.

Output

Your program is to write to standard output. The first line contains one integer, which is the desired minimal number.

Sample Input

5Ab3bd

Sample Output

2

Source

IOI 2000

 

 

题目大意：给出一个字符串，问最少插入多少个字符让这个串变成回文串。

分析：可以直接dp。令f[i,j]表示从i到j最少需要插入多少个字符这个串才变成回文串。

f[i,j]=f[i+1,j-1] (a[i]=a[j])

f[i,j]=min(f[i+1,j]+1,f[i,j-1]+1)。

也可以求原串和反串的最长公共子串，再用长度减去。其实本质是相同的。

 

恶心的是，这题内存限制比较小，都开longint会超内存..

用max函数的话，容易超时...

题目没说，其实是多组测试数据..而且pascal一定要用seekeof= =

c的话，类型用unsigned可以了。

 

codes：

 

var n:longint; a,b:array[1..5000] of char; f:array[0..5000,0..5000] of integer;procedure init; var i:longint; begin readln(n); for i:=1 to n do begin read(a[i]); b[n-i+1]:=a[i]; end; f[0,0]:=0; end;procedure dp; var i,j:longint; begin for i:=1 to n do for j:=1 to n do begin f[i,j]:=-1; if a[i]=b[j] then f[i,j]:=f[i-1,j-1]+1 else if f[i-1,j]>f[i,j-1] then f[i,j]:=f[i-1,j] else f[i,j]:=f[i,j-1] end; writeln(n-f[n,n]); end;begin while not seekeof do begin init; dp; end;end.