POJ - 1159 Palindrome 最长公共子序列

Palindrome Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 61919   Accepted: 21576

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 200

不难想就是求最长回文子序列长度，之前听别人讨论过最长回文子序长度的求法，是字符本身与它的逆序字符的最长公共子序列长度

例如  abbac

也就是求abbac 与cabba的最长公共子序列

写了一发超内存了，改了改才过了

#include<stdio.h> #include<string.h> #include<math.h> #include<string> #include<algorithm> #include<queue> #include<vector> #include<map> #include<set> typedef long long ll; using namespace std; int dp[3][5005]; int main() { int i,j,n; char ch[5005],ch1[5005]; scanf("%d",&n); getchar(); scanf("%s",ch+1); for(i=1;i<=n;i++) ch1[n-i+1]=ch[i]; for(i=1;i<=n;i++) { for(j=1;j<=n;j++) { if(ch[i]==ch1[j]) { dp[1][j]=dp[0][j-1]+1; } else dp[1][j]=max(dp[0][j],dp[1][j-1]); } for(j=1;j<=n;j++) dp[0][j]=dp[1][j]; } printf("%d\n",n-dp[1][n]); return 0; }