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。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 1010;
18 struct DP{
19     int len;
20     string str;
21 };
22 DP dp[maxn][maxn];
23 char sa[maxn],sb[maxn];
24 int main() {
25     while(gets(sa)){
26         int len  = strlen(sa);
27         strcpy(sb,sa);
28         reverse(sb,sb+len);
29         for(int i = 0; i <= len; ++i){
30             dp[0][i].len = 0;
31             dp[0][i].str = "";
32         }
33         for(int i = 1; i <= len; ++i){
34             for(int j = 1; j <= len; ++j){
35                 if(sa[i-1] == sb[j-1]){
36                     dp[i][j].len = dp[i-1][j-1].len+1;
37                     dp[i][j].str = dp[i-1][j-1].str + sa[i-1];
38                 }else if(dp[i-1][j].len > dp[i][j-1].len){
39                     dp[i][j].len = dp[i-1][j].len;
40                     dp[i][j].str = dp[i-1][j].str;
41                 }else if(dp[i-1][j].len < dp[i][j-1].len){
42                     dp[i][j].len = dp[i][j-1].len;
43                     dp[i][j].str = dp[i][j-1].str;
44                 }else{
45                     dp[i][j].len = dp[i-1][j].len;
46                     dp[i][j].str = min(dp[i-1][j].str,dp[i][j-1].str);
47                 }
48             }
49         }
50         string ans = dp[len][len].str;
51         if(dp[len][len].len&1){
52             for(int i = 0; i < dp[len][len].len>>1; ++i)
53                 putchar(ans[i]);
54             for(int i = dp[len][len].len>>1; i >= 0; --i)
55                 putchar(ans[i]);
56             putchar('\n');
57         }else{
58             for(int i = 0; i+1 < dp[len][len].len>>1; ++i)
59                 putchar(ans[i]);
60             for(int i = dp[len][len].len>>1; i >= 0; --i)
61                 putchar(ans[i]);
62             putchar('\n');
63         }
64     }
65     return 0;
66 }
View Code

 

posted @ 2014-10-20 15:14  狂徒归来  阅读(327)  评论(0)  编辑  收藏  举报