Palindromic Subsequence - UVa 11404 dp

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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

aabbaabbcomputerabzlasamhita

Sample Output

aabbaacabaaha

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;    }}

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