f[i][j]表示從i到j染色最少需要多少次

如果a[i]==a[j]，可以從f[i+1][j]，f[i][j-1]，f[i+1][j-1]+1三個狀態轉移。

否則對區間進行分裂，從小區間轉移。

程式碼：

// luogu-judger-enable-o2
//#define LOCAL
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <bits/stdc++.h>
 


#define INF 0x3f
#define ull unsigned long long
#define ll long long
#define FOR(a, b, n) for(int a = b; b >= n ? a >= n : a <= n; b >= n ? a-- : a++)
#define M(a, n) memset(a, n, sizeof(a));
#define S(n) scanf("%s", n)
#define P(n) printf("%d", n)
#define G(n) getline(cin, n)
#define PI acos(-1.0)
 


using namespace std;

const int NR = 60;

int f[NR][NR], n, a[NR];
char b[NR];

inline int read()  {
    int s = 0, w = 1;
    char ch = getchar();
    while(ch <= '0' || ch > '9') {
        if(ch == '-') {
            w = -1;
            ch = getchar();
        }
    }
    while(ch >= '0' &&
 
 ch <= '9') {
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * w;
}

int main() {
    S(b + 1);
    int n = strlen(b + 1);
    int j, l;
    FOR(i, 1, n)
        a[i] = b[i] - 'A' + 1;
    M(f, 0x3f);
    FOR(i, 1, n)
        f[i][i] = 1;
    FOR(l, 2, n) { 
        for(int i = 1; i + l - 1 <= n; i++) {
            j = i + l - 1;
            if(a[i] == a[j]) {
                f[i][j] = min(f[i + 1][j], f[i][j - 1]);
                f[i][j] = min(f[i][j], f[i + 1][j - 1] + 1);
            }
            else {
                FOR(k, i, j - 1) {
                    f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j]);
                }
            }
        }
    }
    P(f[1][n]);
}