动态规划加最少字符使字符串变回文 POJ 1159 short水过！

最新推荐文章于 2022-12-16 21:52:39 发布

最新推荐文章于 2022-12-16 21:52:39 发布 · 184 阅读

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#数据结构与算法

本文介绍了一个算法问题，即如何确定将一个给定字符串转换为回文串所需的最少字符插入次数。通过动态规划的方法，文章提供了一个C语言实现的解决方案，能够有效地解决这一问题。

Palindrome

Time Limit:3000MS		Memory Limit:65536K
Total Submissions:33120		Accepted:11122

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

#include<stdio.h>
#define MAX 5001
short opt[MAX][MAX]={0};//opt[i][j]表示i,j之间至少插入的字符数才能达到对称
//这里数组用int超内存，改成看到网上有的说改成short后，可以水过，果然4000多K水过，不过可以采取滚动数组来优化
int main()
{
    //freopen("input","r",stdin);
    int str_len;
    int i,j;
    char str[MAX];
    scanf("%d",&str_len);
    scanf("%s",str);
    for(i=str_len;i>0;i--)//注意循环的方式，一般使用opt数组的时候，循环下标最小为1，为了防止opt[i-1][j-1],产生数组越界
        for(j=i;j<=str_len;j++)
        {
                if(str[i]==str[j]) opt[i][j]=opt[i+1][j-1];//子结构的方程决定了上述for循环的方式
                else if(opt[i+1][j]>opt[i][j-1]) opt[i][j]=opt[i][j-1]+1;
                else opt[i][j]=opt[i+1][j]+1;

        }
        //printf("%d %d",_i,_j);
    printf("%d",opt[1][str_len]);
}