Palindromic Subsequence - UVa 11404 dp

最新推荐文章于 2018-08-06 18:42:28 发布

提比-我有特殊的AC技巧

最新推荐文章于 2018-08-06 18:42:28 发布

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分类专栏： UVa dp动态规划和递推

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本文介绍了一种使用动态规划解决寻找字符串中最长回文子序列问题的方法，并提供了详细的AC代码实现。该方法不仅计算最长子序列长度，还输出字典序最小的回文子序列。

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Palindromic Subsequence

A Subsequence is a sequence obtained by deleting zero or more characters in a string. A Palindrome is a string which when read from left to right, reads same as when read from right to left. Given a string, find the longest palindromic subsequence. If there are many answers to it, print the one that comes lexicographically earliest.

Constraints

Maximum length of string is 1000.
Each string has characters `a' to `z' only.

Input

Input consists of several strings, each in a separate line. Input is terminated by EOF.

Output

For each line in the input, print the output in a single line.

Sample Input

aabbaabb
computer
abzla
samhita

Sample Output

aabbaa
c
aba
aha

题意：找到最长的非连续回文字串，并输出字典序最小的。

思路：dp[i][j].num表示从i到j的回文串的最大字符数，dp[i][j].str表示这个最小的字符串，每次如果s[i]==s[j]时，那么两端必取这两个字符，否则的话，取dp[i+1][j]和dp[i][j-1]中字符串字典序最小的。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
struct node
{
    int num;
    string str;
}dp[1010][1010];
char s[1010],c;
int n;
int main()
{
    int i,j,k,p;
    while(gets(s+1))
    {
        n=strlen(s+1);
        if(n==0)
        {
            printf("\n");
            continue;
        }
        for(i=1;i<=n;i++)
        {
            dp[i][i].num=1;
            dp[i][i].str=s[i];
            dp[i][i-1].str="";
        }
        for(k=1;k<=n;k++)
           for(i=1;i+k<=n;i++)
           {
               j=i+k;
               dp[i][j].num=max(dp[i+1][j].num,dp[i][j-1].num);
               if(s[i]==s[j])
               {
                   dp[i][j].num=max(dp[i][j].num,dp[i+1][j-1].num+2);
                   dp[i][j].str=s[i]+dp[i+1][j-1].str+s[i];
                   continue;
               }
               if(dp[i][j].num==dp[i+1][j].num && dp[i][j].num==dp[i][j-1].num)
               {
                   dp[i][j].str=min(dp[i+1][j].str,dp[i][j-1].str);
                   continue;
               }
               if(dp[i][j].num==dp[i+1][j].num)
                 dp[i][j].str=dp[i+1][j].str;
               else if(dp[i][j].num==dp[i][j-1].num)
                 dp[i][j].str=dp[i][j-1].str;
           }
        cout<<dp[1][n].str<<endl;
    }
}