OpenJudge/Poj 1159 Palindrome

1.链接地址：

http://bailian.openjudge.cn/practice/1159/

http://poj.org/problem?id=1159

2.题目：

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 49849		Accepted: 17153

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

5

Ab3bd

Sample Output

2

Source

IOI 2000

3.思路：

这题要知道其实最少增加的个数= 字符串总字数 - LCS（最长公共子序列）

所以就转化为求LCS

LCS为典型的dp算法之一，时间复杂度O（n^2），空间复杂度O(n)

!!!这题用string会超时，郁闷。开始对string没好感

4.代码：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 

 5 #define max(a,b) ((a) > (b) ? (a) : (b))

 6 

 7 using namespace std;

 8 

 9 

10 int same(char ch1,char ch2)

11 {

12     if(ch1 == ch2) return 1;

13     else return 0;

14 }

15 

16 int LCS(char *str1,char *str2,int len1,int len2)

17 {

18     int i,j;

19 

20     //if(len1 < len2) {char *str3 = str1;str1 = str2;str2 = str3;}

21 

22     int **dp = new int*[2];

23     for(i = 0; i < 2; ++i) dp[i] = new int[len2 + 1];

24     memset(dp[0],0,sizeof(int) * (len2 + 1));

25     dp[1][0] = 0;

26 

27     

28     for(i = 1; i <= len1; ++i)

29     {

30         for(j = 1; j <= len2; ++j)

31         {

32             dp[i % 2][j] = max(dp[(i - 1) % 2][j],max(dp[i % 2][j - 1],dp[(i - 1) % 2][j - 1] + same(str1[i - 1],str2[j - 1])));

33             //cout<<"dp[" << i << "][" << j << "]=" << dp[i % 2][j] << endl;

34         }

35     }

36     int max = dp[len1 % 2][len2];

37 

38     for(i = 0; i < 2; ++i) delete [] dp[i];

39     delete [] dp;

40 

41     return max;

42 }

43 

44 int main()

45 {

46     int n;

47     cin>>n;

48 

49     char *str1 = new char[n];

50     char *str2 = new char[n];

51 

52     int i;

53     for(i = 0; i < n; ++i)

54     {

55         cin>>str1[i];

56         str2[n - 1 - i] = str1[i];

57     }

58 

59     int lcs_len = LCS(str1,str2,n,n);

60 

61     cout<<(n - lcs_len)<<endl;

62 

63     delete [] str1;

64     delete [] str2;

65     return 0;

66 }