Codeforces Round #311 (Div. 2) E - Ann and Half-Palindrome（字典树+dp）
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 tif
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?
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.
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.
abbabaab 7
abaa
aaaaa 10
aaa
bbaabb 13
bbaabb
By definition, string a = a1a2... an is lexicographically less than string b = b1b2... bm, if either a is a prefix of b and doesn't coincide withb, or there exists such i, that a1 = b1, a2 = b2, ... ai - 1 = bi - 1, ai < bi.
In the first sample half-palindrome substrings are the following strings — a, a, a, a, aa, aba, abaa, abba, abbabaa, b, b, b, b, baab,bab, bb, bbab, bbabaab (the list is given in the lexicographical order).
大致题意：找出s的子串中字典序第k小的“半回文串”，给出半回文串定义是：对于随意i<=|s|/2 有s[i] = s[len-i+1]
数据量是5000
O（n^2）的算法可行
简单暴力的方法就是n^2 dp 出（i,j）的子串是不是半回文串，再把全部子串插入字典树。在dfs遍历出第k小的串
方法二，dp[i][j]统计出以i为始端到i~j的子串是回文串的总数。再把全部子串插入字典树，然后二分出答案子串。
方法一：
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
const int N = 5000+5;
int dp[N][N];
char s[N];
int k;
int n;
struct node
{
node *son[2];
int cnt;
node()
{
memset(son,0,sizeof(son));
cnt = 0;
}
}trie[N*N];
int SZ = 0;
node *root,*cur;
node *createnode()
{
return &trie[SZ++];
}
void Insert(int p)
{
cur = root;
for(int i = p;i <= n;i++)
{
int val = s[i]-'a';
if(cur->son[val] == NULL) cur->son[val] = createnode();
cur = cur->son[val];
cur->cnt += dp[p][i];
}
}
int getsum(node *cur)
{
int sum = 0;
if(cur->son[0]) sum += getsum(cur->son[0]);
if(cur->son[1]) sum += getsum(cur->son[1]);
cur->cnt += sum;
return cur->cnt;
}
vector<char>ans;
bool dfs(node *cur,int &k)
{
if( k <= 0) return true;
if(cur->son[0])
{
ans.push_back('a');
if(cur->son[0]->cnt) k -= cur->son[0]->cnt;
if(dfs(cur->son[0],k)) return true;
ans.pop_back();
}
if(cur->son[1])
{
ans.push_back('b');
if(cur->son[1]->cnt)k -= cur->son[1]->cnt;
if(dfs(cur->son[1],k)) return true;
ans.pop_back();
}
return false;
}
int main()
{
scanf("%s%d",s+1,&k);
n = strlen(s+1);
for(int i =0;i <= n+2;i++)
for(int j = i;j >= 0;j--)
dp[i][j] = 1;
for(int len = 2;len <= n;len++)
for(int l = 1;l+len-1 <= n;l++)
{
int r = l+len-1;
dp[l][r] = (s[l] == s[r] && dp[l+2][r-2]);
}
root = createnode();
REP(i,n)Insert(i);
dfs(root,k);
foreach(i,ans) putchar(*i);
}
方法二：
//GNU C++ Accepted 374 ms 392200 KB
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;
typedef long long ll;
const int N = 5000+5;
int dp[N][N];
char s[N];
int k;
int n;
struct node
{
node *son[2];
int cnt;
node()
{
memset(son,0,sizeof(son));
cnt = 0;
}
}trie[N*N];
int SZ = 0;
node *root,*cur;
node *createnode()
{
return &trie[SZ++];
}
void Insert(int p)
{
cur = root;
for(int i = p;i <= n;i++)
{
int val = s[i]-'a';
if(cur->son[val] == NULL) cur->son[val] = createnode();
cur = cur->son[val];
if(i != p)cur->cnt += dp[p][n]-dp[p][i-1];
else cur->cnt += dp[p][n];
}
}
void query(node*cur,int k) //二分过程
{
if(k <= 0) return ;
if(cur->son[0] && cur->son[0]->cnt >= k)
{
int cnt = cur->son[0]->cnt;
cur = cur->son[0];
if(cur->son[0]) cnt -= cur->son[0]->cnt;
if(cur->son[1]) cnt -= cur->son[1]->cnt;
putchar('a');
query(cur,k-cnt);
}
else if(cur->son[1])
{
if(cur->son[0]) k -= cur->son[0]->cnt;
int cnt = cur->son[1]->cnt;
cur = cur->son[1];
if(cur->son[0]) cnt -= cur->son[0]->cnt;
if(cur->son[1]) cnt -= cur->son[1]->cnt;
putchar('b');
query(cur,k-cnt);
}
}
int main()
{
scanf("%s%d",s+1,&k);
n = strlen(s+1);
for(int i =0;i <= n+2;i++)
for(int j = i;j >= 0;j--)
dp[i][j] = 1;
for(int len = 2;len <= n;len++)
for(int l = 1;l+len-1 <= n;l++)
{
int r = l+len-1;
dp[l][r] = (s[l] == s[r] && dp[l+2][r-2]);
}
REP(i,n)
for(int j = i+1;j <= n;j++) dp[i][j] += dp[i][j-1];
root = createnode();
REP(i,n) Insert(i);
query(root,k);
}
