Baseline Time Limit: 1 second Space Limit: 131072 KB Score: 40
Difficulty: Level 4 Algorithm Questions
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There is a string S, find the minimum number of palindromes that S can be divided into.
For example: abbaabaa, there are many ways to divide.
a|bb|aabaa - 3 palindromes
a|bb|a|aba|a - 5 palindromes
a|b|b|a|a|b|a|a - 8 palindromes
Among them, the first division method has the least number of divisions.
Input
Input string S (length of S <= 5000).
Output
Output the minimum number of divisions.
Input example
father
Output example
3
Idea: dp[i] represents the minimum number of palindrome strings that can be divided into i.
dp[i] = min( dp[i] , dp[j-1]+1); (when ji is a palindrome).
First, use the Manacher algorithm to preprocess to get the P[i] array. When performing dp, take the p[pos] value of the position between i and j. If p[pos]+pos >= i or j==i, then it means that ji This paragraph is a palindrome.
(Manacher's algorithm p[i]+i-1 is the rightmost position of the center of each palindrome, and this position is the '#' sign.)
Code：
#include <bits/stdc++.h>
using namespace std;
const int AX = 1e4+66;
int p[AX];
int dp[AX];
char s[AX];
int main(){
cin >> s ;
int len = strlen(s);
for( int i = len ; i >= 0 ; i-- ){
s[2*i+2] = s[i];
s[2*i+1] = '#';
}
s[0] = '$';
int id = 0 , mx = 0 ;
for( int i = 2 ; i < 2 * len + 1 ; i++ ){
if( p[id] + id > i ){
p[i] = min( p[id] + id - i , p[2*id-i] );
}else p[i] = 1;
while( s[i - p[i]] == s[i+p[i]] ) p[i] ++;
if( mx < p[i] + i ){
mx = p[i] + i;
id = i;
}
}
dp[0] = 1;
for( int i = 1 ; i < len ; i++ ){
dp[i] = i + 1 ;
for( int j = 0 ; j <= i ; j++ ){
int tm_i = 2 * i + 2;
int tm_j = 2 * j + 2;
int pos = ( tm_i + tm_j ) >> 1 ;
if( p[pos] + pos - 1 > tm_i || i == j ){
dp[i] = min( dp[i] , dp[j-1] + 1 );
}
}
}
cout << dp[len-1] << endl;
return 0 ;
}