Palindrome split

Palindrome split

Title description

Given a string, return the minimum number of cuts that cut all str into palindrome strings.

Enter a description:

The input contains a line of string, representing $str(1\le lengt{h}_{str}\le 5000)$。

Output description:

Output an integer, representing the minimum number of cuts to cut all str into palindrome strings.

Example 1

enter

ABA

Output

0

Description

本身是回文串，不需要切割，直接输出0

Example 2

enter

ABCBAEEE

Output

1

Description

切割1次，变为“ABCBA”和“EEE”

Remarks:

Time complexity $O\left(n^2\right)$, additional space complexity$O(n^2)$。

---

answer:

Dynamic programming. Let F[i] denote that str[0...i] needs to be split at least several times before all str[0...i] can be split into palindrome strings. Calculate F[i] from left to right, the process is as follows:

• Suppose the current position is j(j<=i<len), if str[j...i] is a palindrome, then F[i]=F[j-1]+1, which means in str[0...i] On the substring, str[j...i] is already a palindrome string, it is used as a separate part, and the rest uses its minimum number of palindrome divisions, namely F[j-1];
• Let j enumerate from 0 to i, find the minimum value in all cases, that is, F[i]=min{F[j-1]+1(0<=j<=i<len, and str[ j...i] is a palindrome)};
• To quickly determine whether str[j…i] is a palindrome, the process is as follows:
  • Define a two-dimensional array p, $p[j][i]=true$ stands for str[j...i] is a palindrome;
  • 若 $p\left[j\right]\left[i\right]=true$ must be the following three cases:
    • str[j…i] has only one character;
    • str[j…i] has two characters and the two characters are equal;
    • str[j+1…i-1] is a palindrome string, that is, $p[j+1][i-1]=true$ ，且 str[j]==str[i]；
• i goes from left to right, j goes from right to left, so $p[j+1][i-1]$ Must have been calculated.

Code:

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 5005;

char s[N];
bool p[N][N];
int f[N];

int main(void) {
    
    
    scanf("%s", s);
    int n = strlen(s);
    for ( int i = 1; i < n; ++i ) {
    
    
        f[i] = i;
        for ( int j = i; j >= 0; --j ) {
    
    
            if ( s[i] == s[j] && ( i-j<2 || p[j+1][i-1] ) ) {
    
    
                p[j][i] = true;
                if (!j) f[i] = 0;
                else f[i] = min(f[i], f[j - 1] + 1);
            }
        }
    }
    printf("%d\n", f[n - 1]);
    return 0;
}