构造回文

Posted 2022-12-06 taxue505

给定一个字符串s，你可以从中删除一些字符，使得剩下的串是一个回文串。如何删除才能使得回文串最长呢？
输出需要删除的字符个数。

输入描述:

输入数据有多组，每组包含一个字符串s，且保证:1<=s.length<=1000.
  

输出描述:

对于每组数据，输出一个整数，代表最少需要删除的字符个数。

输入例子:

abcda
google

输出例子:

2
2

#include<iostream>
#include<string>
#include<algorithm>

using namespace std;
const int MAX = 1001;
int MaxLen[MAX][MAX]; 

//最长公共子序列，动态规划求法
int maxLen(string s1, string s2)

    int length1 = s1.size();
    int length2 = s2.size();
    for (int i = 0; i < length1; ++i)
        MaxLen[i][0] = 0;
    for (int i = 0; i < length2; ++i)
        MaxLen[0][i] = 0;
     
    for (int i = 1; i <= length1; ++i)
    
        for (int j = 1; j <= length2; ++j)
        
            if (s1[i-1] == s2[j-1])
                MaxLen[i][j] = MaxLen[i-1][j - 1] + 1;
            
            else
            
                MaxLen[i][j] = max(MaxLen[i - 1][j], MaxLen[i][j - 1]);
            
        
    
 
    return MaxLen[length1][length2];

 
int main()
    string s;
    while (cin >> s)
        int length = s.size();
        if (length == 1)
            cout << 1 << endl;
            continue;
        
        //利用回文串的特点
        string s2 = s;
        reverse(s2.begin(),s2.end());
        int max_length = maxLen(s, s2);
        cout << length - max_length << endl;
    
    return 0;