HDU 1513 oldest sequence

Palindrome

Time limit:4000/2000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3641 Accepted Submission (s): 1252

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

Sample INPUT5AB3BD

Sample Output2

Sourceioi 2000 title meaning: to a string of length n, ask at least how many characters are added to make the string a palindrome. Idea: Turn the original string upside down, two strings to the eldest son sequence m, then N-m is the number of characters added. The longest subsequence length means the fewest number of characters added. Because N is too large, you need to use a scrolling array.  DP[I&AMP;1][J] means dp[i][j]; dp[! ( i&1)][j] = dp[i-1][j] Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <iostream>5#include <vector>6#include <queue>7#include <cmath>8#include <Set>9 using namespacestd;Ten  One #defineN 5005 A  - intMaxintXintY) {returnX>y?x:y;} - intMinintXintY) {returnX<y?x:y;} the intAbsintXintY) {returnx<0?-x:x;} -  -  - intdp[2][n]; + intN; - CharS[n]; +  A Main () at { -     intI, J, K; -      while(SCANF ("%d", &n) = =1){ -Memset (DP,0,sizeof(DP)); -scanf"%s", s+1); -         CharS1[n]; in          for(i=n;i>=1; i--) s1[i]=s[n-i+1]; -         intMaxh=0; to          for(i=1; i<=n;i++){ +memset (dp[i&1],0,sizeof(dp[i&1])); -              for(j=1; j<=n;j++){ the                 if(s[i]==S1[j]) { *dp[i&1][j]=max (dp[! ( i&1)][j-1]+1,dp[i&1][j]); $                 }Panax Notoginseng                 Elsedp[i&1][j]=max (dp[! ( i&1)][j],dp[i&1][j-1]); -Maxh=max (maxh,dp[i&1][j]); the             } +         } Aprintf"%d\n", N-maxh); the     } +}

HDU 1513 oldest sequence