poj1159——Palindrome（组成回文串的最少字符数，dp）

最小插入构造回文串

最新推荐文章于 2019-01-24 18:08:04 发布

原创最新推荐文章于 2019-01-24 18:08:04 发布 · 678 阅读

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#string #poj #as #dp

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本文介绍了一种算法，用于确定将任意字符串转换为回文串所需的最少字符插入数量。通过动态规划方法，该算法有效地解决了问题，并提供了一个C++实现示例。

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

普通的方法是求出这个串与它的逆序串的最大公共子序列长度，然后用串长减去这个长度
还有一种是用dp[i][j]表示从i到j需要加上的最少字符数
if(a[i]==a[j])
dp[i][j]=dp[i+1][j-1]
else
dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#define MAXN 5010
using namespace std;
char a[MAXN];
short dp[MAXN][MAXN]; //坑点，short才能ac
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        scanf("%s",a+1);
        for(i=n; i>=1; --i)
        {
            dp[i][i]=0;
            for(j=i+1; j<=n; ++j)
            {
                if(a[i]==a[j])
                    dp[i][j]=dp[i+1][j-1];
                else
                    dp[i][j]=min(dp[i+1][j],dp[i][j-1])+1;
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}