poj1159 Palindrome DP,lcs

最新推荐文章于 2022-05-16 16:41:42 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所需的最少字符插入次数。通过寻找最长对称子序列，该算法有效地解决了问题，并提供了一个实际的C++实现示例。

Palindrome

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 44186		Accepted: 15050

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

Source

求最长公共子序列，第一想到的是用二维数组标记，结果内存超了，题目要求65525k ,还要求n <= 5000,开5000*5000的int 数组果断超啊。。。

同学告诉我可以把4个字节的int 改成无符号短整型 2个字节的

unsigned short int a[5001][5001];

结果内存险过，呵呵。

这题最好的方法是用滚动数组只需a[2][5001];即可

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
 int a[2][5001];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int n,i,j,ii,jj,l;
    while(scanf("%d",&n)!=EOF)
    {
        char  s[5010],rs[5010];
        cin>>s;
        l=strlen(s);
        for(i = 0 ; i < l ; i++)
            rs[i]=s[l-1-i];
        memset(a,0,sizeof(a));
        for(i=0; i<l; ++i)
            for(j=0; j<l; ++j)
            {
                if(s[i]==rs[j])
                    a[(i+1)%2][j+1]=a[i%2][j]+1;
                else
                    a[(i+1)%2][j+1]=max(a[i%2][j+1],a[(i+1)%2][j]);

            }
//        for(int i = 0 ; i < l ; i++)
//            printf("%d ",a[0][l]);
//        puts("");
//        for(int i = 0 ; i < l ; i++)
//            printf("%d ",a[1][l]);
//        puts("");
        printf("%d\n",l-a[l%2][l]);
    }
return 0;
}