UVa11404Palindromic Subsequence（最大回文串，区间DP）

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

题意：

给定一个字符串s，对s进行删除操作，使得剩下的子串是回文字符串，输出最长的字符串，当有多个相同长度的就输出字典序最小的。

思路：

由于要输出字符串，所以在状态转移过程中要保存下字符串，用C++的string就方便很多，然后就是和找最长回文的方法一样了。

定义结构体保存长度，以及字符串，分别用len, s 表示

状态的转移方程为，如果头尾相同，dp[i][j].len = dp[i + 1][j - 1].len + 2（长度加上首尾，所以增加2)；如果首尾不同，那么回文长度不增加 dp[i][j].len = max(dp[i + 1][j].len, dp[i][j - 1].len);

如果长度相同  dp[i + 1][j].len, == dp[i][j - 1].len，那么就要比较子串的字典序，取字典序小的，

也就是判断 dp[i + 1][j].s,  dp[i][j - 1].s。

<span style="font-size:18px;">#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN  = 1010;
struct str
{
    int len;
    string s;
} dp[MAXN][MAXN];
char s[MAXN];

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    while(scanf("%s", s+1) != EOF)
    {
        int len = strlen(s+1);
        for(int i = 1; i <= len; i++)
        {   //初始化
            dp[i][i].len = 1;
            dp[i][i].s = s[i];
        }
        for(int k = 2; k <= len; k++) //控制区间大小
        {
            for(int i = 1, j = k; j <= len; i++, j++) //正推
            //for(int i = len-k+1, j = len; i >= 1; i--, j--) //逆推
            {
                if(s[i] == s[j])
                {
                    dp[i][j].len = dp[i+1][j-1].len+2;
                    dp[i][j].s = s[i]+dp[i+1][j-1].s+s[j];
                }
                else
                {
                    if(dp[i][j-1].len > dp[i+1][j].len ||
                            (dp[i][j-1].len==dp[i+1][j].len&&dp[i][j-1].s<dp[i+1][j].s))
                    {   //当 [i, j-1] 的长度大于 [i+1, j] 的长度，或者二者长度相等并且
                        //[i, j-1] 的字典序小于 [i+1, j] 的字典序，则选择 [i, j-1]，否则选择后者
                        dp[i][j].len = dp[i][j-1].len;
                        dp[i][j].s = dp[i][j-1].s;
                    }
                    else
                    {
                        dp[i][j].len = dp[i+1][j].len;
                        dp[i][j].s = dp[i+1][j].s;
                    }
                }
            }
        }
        cout<<dp[1][len].s<<endl;
    }
    return 0;
}
</span>