UVA 11404 Palindromic Subsequence

Palindromic Subsequence

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 11404
64-bit integer IO format: %lld      Java class name: Main

 

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
 
解题：求最长的且字典序最小的回文子序列。把原串逆转，然后与原串求LCS。LCS的的前半部分一定要求的回文序列的前半部分，但是后半部分可能不是。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 struct DP{
     int len;
     string str;
 };
 DP dp[maxn][maxn];
 char sa[maxn],sb[maxn];
 int main() {
     while(gets(sa)){
         int len  = strlen(sa);
         strcpy(sb,sa);
         reverse(sb,sb+len);
         for(int i = ; i <= len; ++i){
             dp[][i].len = ;
             dp[][i].str = "";
         }
         for(int i = ; i <= len; ++i){
             for(int j = ; j <= len; ++j){
                 if(sa[i-] == sb[j-]){
                     dp[i][j].len = dp[i-][j-].len+;
                     dp[i][j].str = dp[i-][j-].str + sa[i-];
                 }else if(dp[i-][j].len > dp[i][j-].len){
                     dp[i][j].len = dp[i-][j].len;
                     dp[i][j].str = dp[i-][j].str;
                 }else if(dp[i-][j].len < dp[i][j-].len){
                     dp[i][j].len = dp[i][j-].len;
                     dp[i][j].str = dp[i][j-].str;
                 }else{
                     dp[i][j].len = dp[i-][j].len;
                     dp[i][j].str = min(dp[i-][j].str,dp[i][j-].str);
                 }
             }
         }
         string ans = dp[len][len].str;
         if(dp[len][len].len&){
             for(int i = ; i < dp[len][len].len>>; ++i)
                 putchar(ans[i]);
             for(int i = dp[len][len].len>>; i >= ; --i)
                 putchar(ans[i]);
             putchar('\n');
         }else{
             for(int i = ; i+ < dp[len][len].len>>; ++i)
                 putchar(ans[i]);
             for(int i = dp[len][len].len>>; i >= ; --i)
                 putchar(ans[i]);
             putchar('\n');
         }
     }
     return ;
 }