题目链接

题意：



题解：













AC代码

/*
    Author : zzugzx
    Lang : C++
    Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;

#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;

const double eps = 1e-6;
const double PI = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 410;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};

int dp[55][55];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
//  freopen("in.txt", "r", stdin);
//  freopen("out.txt", "w", stdout);
    string s;
    cin >> s;
    int n = s.length();
    s = '.' + s;
    mem(dp, inf);
    for (int i = 1; i <= n; i++) dp[i][i] = 1;
    for (int len = 2; len <= n; len++)
        for (int l = 1, r = l + len - 1; r <= n; l++, r++) {
            if (s[l] == s[r])
                dp[l][r] = min(dp[l + 1][r], dp[l][r - 1]);
            else
                for (int k = l; k < r; k++)
                    dp[l][r] = min(dp[l][r], dp[l][k] + dp[k + 1][r]);
        }
    cout << dp[1][n];
    return 0;
}