UVa11404Palindromic Subsequence最大回文串，区间DP

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1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，所以在状态转移进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1样了。
定义结构体保存长度，和字符串，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，那末回文长度不增加 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，那末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(⑴.0); const double e = 2.718281828459; const double eps = 1e⑻; 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][j⑴].len+2; dp[i][j].s = s[i]+dp[i+1][j⑴].s+s[j]; } else { if(dp[i][j⑴].len > dp[i+1][j].len || (dp[i][j⑴].len==dp[i+1][j].len&&dp[i][j⑴].s<dp[i+1][j].s)) { //当 [i,j⑴] 的长度大于 [i+1,j] 的长度，或2者长度相等并且 //[i,j⑴] 的字典序小于 [i+1,j] 的字典序，则选择 [i,j⑴]，否则选择后者 dp[i][j].len = dp[i][j⑴].len; dp[i][j].s = dp[i][j⑴].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>