UVA 11404 - Palindromic Subsequence(dp)

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 

aabbaabb
computer
abzla
samhita

Sample Output 

aabbaa
c
aba
aha

题意：给定一个字符串，求删除一些字符后，最长的回文字符串，并且要求字典序最小的。

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string就方便很多，然后就是和找最长回文的方法一样了。状态的转移方程为，如果头尾相同，dp[i][j] = dp[i + 1][j - 1] + 2（长度多上首尾多2)如果首尾不同，那么回文长度不增加dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);

代码：

#include <stdio.h>
#include <string.h>
#include <string>
#include <iostream>
using namespace std;

const int N = 1005;

struct State {
	int value;
	string str;
	bool operator > (State &a) {
		if (value != a.value)
			return value > a.value;
		return str < a.str;
	}
} dp[N][N];

int n;
char s[N];

int main() {
	while (~scanf("%s", s + 1)) {
		n = strlen(s + 1);
		for (int i = n; i >= 1; i--)
			for (int j = i; j <= n; j++) {
				if (s[i] == s[j]) {
					if (i == j) {
						dp[i][j].value = 1;
						dp[i][j].str = s[i];
					}
					else {
						dp[i][j].value = dp[i + 1][j - 1].value + 2;
						dp[i][j].str = s[i] + dp[i + 1][j - 1].str + s[j];
					}
				}
				else {
					if (dp[i + 1][j] > dp[i][j - 1]) {
						dp[i][j].value = dp[i + 1][j].value;
						dp[i][j].str = dp[i + 1][j].str;
					}
					else {
						dp[i][j].value = dp[i][j - 1].value;
						dp[i][j].str = dp[i][j - 1].str;
					}
				}
			}
		cout << dp[1][n].str << endl;
	}
	return 0;
}