Palindrome(最长公共子序列)

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 48526		Accepted: 16674

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

题意：给一个字符串的长度和这个字符串s[]，问将这个字符串变成回文串至少要加几个字符；

思路：将这个字符串逆置s1[],求出s[]和s1[]的最长公共子序列的长度len，然后n-len即是要求的长度；

 1 #include<stdio.h>

 2 #include<string.h>

 3 short dp[5001][5001];

 4 int main()

 5 {

 6     int i,j,n;

 7     char s[5010],s1[5010];

 8     memset(dp,0,sizeof(dp));

 9 

10     scanf("%d",&n);

11     scanf("%s",s);

12 

13     for(int i = 0; i < n; i++)

14         s1[i] = s[n-1-i];

15     s1[n] = '\0';

16 

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

18     {

19         for(j = 0; j < n; j++)

20         {

21             if(s[i] == s1[j])

22                 dp[i+1][j+1] = dp[i][j]+1;

23             else

24             {

25                 if(dp[i+1][j] > dp[i][j+1])

26                     dp[i+1][j+1] = dp[i+1][j];

27                 else dp[i+1][j+1] = dp[i][j+1];

28             }

29         }

30     }

31     int ans = n-dp[n][n];

32     printf("%d\n",ans);

33     return 0;

34 }

View Code