Codeforces Round #311 (Div. 2) E. Ann and Half-Palindrome 字典树/半回文串

E. Ann and Half-Palindrome

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/557/problem/E

Description

Tomorrow Ann takes the hardest exam of programming where she should get an excellent mark.

On the last theoretical class the teacher introduced the notion of a half-palindrome.

String t is a half-palindrome, if for all the odd positions i () the following condition is held: ti = t|t| - i + 1, where |t| is the length of string t if positions are indexed from 1. For example, strings "abaa", "a", "bb", "abbbaa" are half-palindromes and strings "ab", "bba" and "aaabaa" are not.

Ann knows that on the exam she will get string s, consisting only of letters a and b, and number k. To get an excellent mark she has to find the k-th in the lexicographical order string among all substrings of s that are half-palyndromes. Note that each substring in this order is considered as many times as many times it occurs in s.

The teachers guarantees that the given number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Can you cope with this problem?

Input

The first line of the input contains string s (1 ≤ |s| ≤ 5000), consisting only of characters 'a' and 'b', where |s| is the length of string s.

The second line contains a positive integer k —  the lexicographical number of the requested string among all the half-palindrome substrings of the given string s. The strings are numbered starting from one.

It is guaranteed that number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Output

Print a substring of the given string that is the k-th in the lexicographical order of all substrings of the given string that are half-palindromes.

Sample Input

abbabaab
7

Sample Output

abaa

HINT

 

题意

定义了半回文串，即奇数位置上的数是回文的

给你个字符串，在他的子串中，让你找到第k大的半回文子串

题解：

先暴力出回文串的子串

然后强行插入字典树，再强行dfs一下

超级大暴力就好了

代码来源于http://codeforces.com/contest/557/submission/11866823

代码

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 5005
#define mod 10007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************


struct trie
{
    int cnt,ch[2];
};
int ok[maxn][maxn];
trie T[maxn*(maxn+1)/2];
trie null;
char res[maxn];
string s;
int k,ts,m=-1,root,n;
int newnode()
{
    T[++ts]=null;
    return ts;
}
void add(int i)
{
    int cur=root;
    for(int j=i;j<n;j++)
    {
        if(T[cur].ch[s[j]-'a']==-1)
            T[cur].ch[s[j]-'a']=newnode();
        cur=T[cur].ch[s[j]-'a'];
        if(ok[i][j])
            T[cur].cnt++;
    }
}
void dfs(int cur) 
{
    k-=T[cur].cnt;
    if (k<=0) {
        printf("%s",res);
        exit(0);
    }
    if (T[cur].ch[0]!=-1) {
        res[++m]='a';
        dfs(T[cur].ch[0]);
        res[m]=0;m--;
    }
    if (T[cur].ch[1]!=-1) {
        res[++m]='b';
        dfs(T[cur].ch[1]);
        res[m]=0;m--;
    }
}
int main()
{
    cin>>s>>k;
    n=s.size();
    null.cnt=0,null.ch[0]=null.ch[1]=-1;
    root=newnode();
    for(int len=1;len<=n;len++)
    {
        for(int i=0;i<=n-len;i++)
        {
            int j=i+len-1;
            if(j-i<=3)
                ok[i][j]=(s[i]==s[j]);
            else
                ok[i][j]=(s[i]==s[j])&&ok[i+2][j-2];
        }
    }
    for(int i=0;i<n;i++)
        add(i);
    dfs(root);
}