专题测试三 贪心与动态规划 H - Palindromic Subsequence

1. 题目
   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
2. 思路
   寻找不连续的回文字符串，dp[i][j]表示从i到j的最长回文串字符数，str表示字符串，如果s[i]==s[j]，就在两端加上这个字符，否则取dp[i+1][j]和dp[i][j-1]中字符串最长、字典序最小的。
3. 代码
   

   #include<cstdio>
   #include<iostream>
   #include<string>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   
   int dp[1050][1050];
   string str[1050][1050];
   
   char s[1050];
   int n,i,j,k,p;
   
   int main()
   {
       while(scanf("%s",s+1)!=EOF)
       {
   		n=strlen(s+1);
           for(i=1;i<=n;i++)
           {
               dp[i][i]=1;
               str[i][i]=s[i];
           }
           for(k=1;k<=n;k++)
   			for(i=1;i+k<=n;i++)
   			{
   				j=i+k;
   				dp[i][j]=max(dp[i+1][j], dp[i][j-1]);
   				if(s[i] == s[j])
   				{
   					dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
   					str[i][j]=s[i]+str[i+1][j-1]+s[j];
   				}
   				else if(dp[i+1][j] == dp[i][j-1])
   					str[i][j]=min(str[i+1][j],str[i][j-1]);
   				else if(dp[i][j]   == dp[i+1][j])
   					str[i][j]=str[i+1][j];
   				else if(dp[i][j]   == dp[i][j-1])
   					str[i][j]=str[i][j-1];
              }
   		cout<<str[1][n]<<endl;
       }
   }