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POJ 1159 Palindrome (LCS)

2019-11-11 01:18:16
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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

題意

給出一個字符串,問最少添加幾個字符才能構成一個回文串。

思路

對于這種類型的題目是有一個公式的啦!

添加的最少字符個數 = 字符串長度 - (字符串正序與逆序的最長公共子序列長度)

但是,對于這道題,僅僅知道這個公式可不行哦~

題目中有說明字符串長度最大為5000,但是dp數組開到 5000?5000 的話會內存超限。

有兩種解決方案:

使用short類型滾動數組

AC 代碼

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<vector>#include<queue>#include<map>#define MAXX 5100char c[MAXX],d[MAXX];int dp[2][MAXX],n;void lcs(){ for(int i=1; i<=n; i++) for(int j=1; j<=n; j++) { if(c[i-1]==d[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)]); } cout<<n-dp[n%2][n]<<endl; //字符串長度-最長公共子序列}int main(){ while(cin>>n>>c) { memset(dp,0,sizeof(dp)); for(int i=n-1; i>=0; i--) d[n-i-1]=c[i]; lcs(); } return 0;}
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