UVA 11404 Palindromic Subsequence (去除字符DP，4级)

E - Palindromic Subsequence

Crawling in process...

Crawling failed

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

​​Submit​​​ ​​​Status​​

System Crawler (2013-05-30) acmparand (2013-08-02)

Description

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

去除

去除最少的字符使其是回文串，字典序最小。直接记录每次最优更新用string

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=1005;
char s[mm],rs[mm];
class Node
{
  public:
  int len;string s;
  bool operator<(const Node&x)const
  {
    if(len!=x.len)return len<x.len;
    return s>x.s;
  }
}dp[mm][mm];
void LCS(int len)
{
  FOR(i,0,len)
  {
    dp[i][0].len=0;dp[i][0].s.clear();
    dp[0][i].len=0;dp[0][i].s.clear();
  }
  FOR(i,1,len)FOR(j,1,len)
  {
    if(s[i]==rs[j])//lcs
    {
      dp[i][j].len=dp[i-1][j-1].len+1;
      dp[i][j].s=dp[i-1][j-1].s+s[i];
    }
    else
    {
      dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
    }
  }
}
string revrse(string s)
{
  string tmp="";
  for(int i=s.length()-1;i>=0;--i)
  tmp+=s[i];
  return tmp;
}
string getans(int len)
{
  Node ans;
  ans.s.clear();ans.len=0;
  FOR(i,1,len-1)
  {
    ans=max(ans,dp[i][len-i]);
  }
  ans.len+=ans.len;
  ans.s+=revrse(ans.s);
  Node tmp;
  FOR(i,1,len)
  { tmp.len=2*dp[i-1][len-i].len+1;
    tmp.s=dp[i-1][len-i].s+s[i]+revrse(dp[i-1][len-i].s);
    ans=max(ans,tmp);
  }
  return ans.s;
}
int main()
{
  while(cin>>s+1)
  {int len=strlen(s+1);
   FOR(i,1,len)rs[i]=s[len-i+1];
    LCS(len);
    cout<<getans(len)<<"\n";
  }
  return 0;
}