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

最新推荐文章于 2022-09-19 09:23:25 发布

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最新推荐文章于 2022-09-19 09:23:25 发布

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原文链接：http://www.cnblogs.com/nealgavin/p/3797525.html

本文介绍了一个算法，用于在给定字符串中找到最长的回文子序列，并确保该序列在字典序上是最小的。算法通过记录每次更新的最优解来实现这一目标。

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E - Palindromic Subsequence

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status

Appoint description: 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;
}

转载于:https://www.cnblogs.com/nealgavin/p/3797525.html