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 }

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