HDU-1513 Palindrome 【LCS+滚动数组】

最新推荐文章于 2018-05-16 20:51:49 发布

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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。通过计算原始字符串与其反转字符串之间的最长公共子序列(LCS)，可以得出所需插入的最小字符数。

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Palindrome

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5241 Accepted Submission(s): 1796

Problem 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

题意：给一个串，问最少添加多少个字符能使他成为回文串（正读反读一样）；
思路：要构成回文串，那就看正串和反串相比有多少个元素不相同，只要又不相同的就添加上去，让不相同的最少就让相同的最多，就是求正串反串的LCS长度；
失误：刚开始认为一个是回文串那就是中心对称 1-N/2 与 N-N/2+1 元素相同，我就求这两个串的LCS了，怎么交怎么不对，第二天举了个例子才发现是不相同的如asd 44d 按照我的思路就把两个4分到了同一个串，肯定不是最优解，还是举个例子验证思路比较好呀！
代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN=1e4+10;
char str[MAXN],sl[MAXN];
int dp[2][MAXN];//滚动数组 i只取两种状态就可以用 和01背包答空间优化类似

int main()
{
    int N,i,j;
    while(~scanf("%d",&N)) {
        scanf("%s",str+1); 
        for(i=N;i>=1;--i) sl[N-i+1]=str[i];
        memset(dp,0,sizeof(dp));
        for(i=1;i<=N;++i)//从1开始防止i-1越界 
        {
            for(j=1;j<=N;++j)
            {
                if(str[i]==sl[j]) dp[i%2][j]=dp[(i-1)%2][j-1]+1;//容易写错 
                else dp[i%2][j]=max(dp[(i-1)%2][j],dp[i%2][j-1]);
            }
        }
        printf("%d\n",N-dp[N%2][N]);
    }
    return 0;
}