标签:codeforces
给定半回文子串的定义,现给你一个串S和一个整数K,输出S所有子串中且是半回文排名第K的子串,半回文子串按照字典序升序顺序。(len(S) <= 5000)
#include<bits\stdc++.h>
#define MAX_ASCII 2
using namespace std;
const int N = 5e+3 + 7;
bool dp[N][N];
char str[N];
int dict[N * N][MAX_ASCII + 1], cnt = 1;
//字典树插入
void Insert(const char *s, int L, int R)
{
int v = 0, i = L;
while (L <= R)
{
if (dict[v][s[L] - 'a'])
v = dict[v][s[L++] - 'a'];
else
v = dict[v][s[L++] - 'a'] = cnt++;
if (dp[i][L - 1])
++dict[v][2];
}
}
//先序遍历查询第K大值
bool Rank(int v, vector<char> &path, int &k)
{
k -= dict[v][2];
if (k <= 0)
{
for (auto x : path)
cout << x;
cout << endl;
return true;
}
for (int i = 0; i < 2; ++i)
{
if (dict[v][i])
{
path.push_back((char)(i + 'a'));
if (Rank(dict[v][i], path, k))
return true;
path.pop_back();
}
}
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(NULL);
int k, len;
cin >> str >> k;
len = strlen(str);
for (int i = len - 1; i >= 0; --i)
for (int j = len - 1; j >= i; --j)
dp[i][j] = i <= j - 4 ? (str[i] == str[j] && dp[i + 2][j - 2]) : str[i] == str[j];
for (int i = 0; i < len; ++i)
{
int j = len - 1;
for (; j >= i && !dp[i][j]; --j);
Insert(str, i, j);
}
vector<char> path;
Rank(0, path, k);
return 0;
}
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Codeforces 311(div 2):E. Ann and Half-Palindrome
标签:codeforces
原文地址:http://blog.csdn.net/dream_you_to_life/article/details/46802147