HDU1513(最大公共子串LCS+DP)

Palindrome

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2262    Accepted Submission(s): 784

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
 
 
 
 
 
 
//仔细想想，确实是求最大公共子串LCS，将题意将其转化为求原字符串与其反字符串的LCS，然后所开数组不宜开【5000】【5000】,这样会超内存，血淋淋的教训啊！为了避免可以开个滚动数组.
#include<iostream>#include<string.h>using namespace std;#define MAX 5000+10int  max(int a,int b){ return a<b?b:a;}char ystr[MAX];char xstr[MAX];int dp[2][MAX];int main(){ int n,i,j,k; int ans; while(scanf("%d",&n)!=EOF) {  scanf("%s",ystr);  strcpy(xstr,ystr);  strrev(xstr);  memset(dp,0,sizeof(dp)); /* for(i=0;i<n;i++)  {   if(ystr[0]==xstr[i])    dp[0][i]=1;  }*/   for(i=1;i<=n;i++)  {   for(j=1;j<=n;j++)   {    if(ystr[i-1]==xstr[j-1])     dp[i%2][j]=dp[(i-1)%2][j-1]+1;    else     {          dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);    }   }  } // printf("ystr=%s    xstr=%s     dp=%d     %d   %d\n",ystr,xstr,dp[n][n],strlen(ystr),strlen(xstr));      ans=n-dp[n%2][n];   printf("%d\n",ans); } return 0;}