POJ 1159 Palindrome

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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

5Ab3bd

Sample Output

2

簡單的 區間動規，主要是容易MLE。

AC代碼如下：

#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時不用補字母        }        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])//此時[i,j]需要補字母數等於[i+1,j-1]中所需補的字母                    dp[i][j]=dp[i+1][j-1];                else                {                    dp[i][j]=min(dp[i][j-1],dp[i+1][j])+1;//此時需要找兩個縮小區間的最小值                }            }        }        cout<<dp[1][n]<<endl;    }    return 0;}