首页 > 代码库 > POJ 1159 Palindrome

POJ 1159 Palindrome

2024-07-11 04:04:11   225人阅读

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. 

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，主要是容易MLE。


AC代码如下：

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
char a[5005];
short dp[5005][5005];//用short就可以不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]需要补字母数等于[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;//此时需要找两个缩小区间的最小值
                }

            }
        }
        cout<<dp[1][n]<<endl;
    }
    return 0;
}



联系
我们