HDU 1513 palindrome "LCS"

Test instructions: Give a string s, ask at least how many letters to make it into a palindrome

Solution: Find the length of the longest common subsequence after the string is flipped, and subtract the length of the longest common subsequence by the length of the string.

Reflection: Because the title gives the range of n is 3<=n<=5000, so the dp[][] array if open to dp[5005][5005], will be super memory, this time should be used to optimize the scrolling array

Detailed introduction to scrolling arrays http://blog.csdn.net/niushuai666/article/details/6677982

Palindrome

Time limit:4000/2000 MS (java/others) Memory limit:65536/32768 K (java/others) total submission (s): 3301 Accepted Submission (s): 1140

Problem Descriptiona Palindrome is a symmetrical string, which is, a string read identically from the left to the right as well as From the right to the left. You-to-write a program which, given a string, determines the minimal number of characters to being inserted into the Stri Ng in order to obtain a palindrome.
As an example, by inserting 2 characters, the string "ab3bd" can is transformed into a palindrome ("Dab3bad" or "Adb3bda" ). However, inserting fewer than 2 characters does not produce a palindrome.  inputyour program was 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 is-to-be considered distinct. outputyour, is-to-write-to-standard output. The first line contains one integer, which is the desired minimal number. sample input5ab3bd sample output2&nbsp ;

#include <stdio.h> #include <string.h>char s[5005],w[5005];int dp[2][5005];int max (int a,int b) {if (a>b) return A;elsereturn B;} int main () {int n,i,j,x,y;while (scanf ("%d", &n) ==1) {scanf ("%s", &s); for (i=0;i<n;i++) w[i]=s[n-1-i];w[i]= ' Memset (Dp,0,sizeof (DP)); for (i=1;i<=n;i++) {for (j=1;j<=n;j++) {   x=i%2;   Y=1-x;  if (S[i-1]==w[j-1])  dp[x][j]=dp[y][j-1]+1;  else  Dp[x][j]=max (dp[y][j],dp[x][j-1]);}    } printf ("%d\n", N-dp[n%2][n]);}}

　　

HDU 1513 palindrome "LCS"