hdu 1513 Palindrome (滚动数组处理回文串)

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6433    Accepted Submission(s): 2127

Problem 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

5Ab3bd

 

Sample Output

2

 

Source

IOI 2000

 

Recommend

linle

将字符串反转存储在新的字符串中。

两个字符串求最长公共子序列。。

然后len- LCM 就是可进行的最少操作数。

因为5000^2的数组爆内存，而且每次的操作只和上一行有关所以就可以滚动数组。

具体操作看代码。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char M[5003];char S[5003];int dp[2][5003];int main(){    int len;    while(~scanf("%d",&len))    {        scanf("%s",M+1);       // printf("---%s\n",M+1);        for(int i=1;i<=len;i++)        {            S[len-i+1]=M[i];        }       // printf("---%s\n",S+1);        memset(dp,0,sizeof(dp));        for(int i=1;i<=len;i++)        {            for(int j=1;j<=len;j++)            {                dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);                if(M[i]==S[j])                {                    dp[i%2][j]=max(dp[i%2][j],dp[(i-1)%2][j-1]+1);                }            }        }        printf("%d\n",len-dp[len%2][len]);    }}