Palindrome
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 51518 | Accepted: 17733 |
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.
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
简单的 区间动规，主要是easyMLE。
AC代码例如以下：
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
char a[5005];
short dp[5005][5005];//用short就能够不MLE了
int min(int a,int b)
{
return a<b?a:b;
}
int main()
{
int n;
int ans;
int i,j;
while(cin>>n)
{
memset(dp,0,sizeof dp);
cin>>a+1;
//cout<<a+1;
for(i=1;i<=n;i++)
{
dp[i][i]=0;//i==j时不用补字母
}
int len;
for(len=2;len<=n;len++)
{
for(i=1;i<=n-len+1;i++)
{
j=i+len-1;
if(a[i]==a[j])//此时[i,j]须要补字母数等于[i+1,j-1]中所需补的字母
dp[i][j]=dp[i+1][j-1];
else
{
dp[i][j]=min(dp[i][j-1],dp[i+1][j])+1;//此时须要找两个缩小区间的最小值
}
}
}
cout<<dp[1][n]<<endl;
}
return 0;
}