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 = 1010;
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 = 0; i <= len; ++i){
            dp[0][i].len = 0;
            dp[0][i].str = "";
        }
        for(int i = 1; i <= len; ++i){
            for(int j = 1; j <= len; ++j){
                if(sa[i-1] == sb[j-1]){
                    dp[i][j].len = dp[i-1][j-1].len+1;
                    dp[i][j].str = dp[i-1][j-1].str + sa[i-1];
                }else if(dp[i-1][j].len > dp[i][j-1].len){
                    dp[i][j].len = dp[i-1][j].len;
                    dp[i][j].str = dp[i-1][j].str;
                }else if(dp[i-1][j].len < dp[i][j-1].len){
                    dp[i][j].len = dp[i][j-1].len;
                    dp[i][j].str = dp[i][j-1].str;
                }else{
                    dp[i][j].len = dp[i-1][j].len;
                    dp[i][j].str = min(dp[i-1][j].str,dp[i][j-1].str);
                }
            }
        }
        string ans = dp[len][len].str;
        if(dp[len][len].len&1){
            for(int i = 0; i < dp[len][len].len>>1; ++i)
                putchar(ans[i]);
            for(int i = dp[len][len].len>>1; i >= 0; --i)
                putchar(ans[i]);
            putchar('
');
        }else{
            for(int i = 0; i+1 < dp[len][len].len>>1; ++i)
                putchar(ans[i]);
            for(int i = dp[len][len].len>>1; i >= 0; --i)
                putchar(ans[i]);
            putchar('
');
        }
    }
    return 0;
}