删删(抵消)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}

void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    string a = s;
    reverse(all(a));
    if (a == s) {
        puts("0");
    }
    else {
        int ans = 0x3f3f3f3f;
        reverse(all(a));
        for (char ch = 'a'; ch <= 'z'; ch ++) {
            s = a;
            int cnt = 0;
            
            while (s.size() > 1) {
                int l = 0, r = s.size() - 1;
                
                if (s[l] != s[r]) {
                    if (s[l] != ch && s[r] != ch) {
                        break;
                    }
                    else if (s[l] == ch) {
                        cnt ++;
                        s.erase(s.begin());
                    }
                    else {
                        cnt ++;
                        s.erase(s.size() - 1 + s.begin());
                    }
                }
                else {
                    s.erase(s.begin());
                    s.erase(s.begin() + s.size() - 1);
                }
            }
            if (s.size() <= 1) {
                ans = min (cnt, ans);
            }
        }
        if (ans == 0x3f3f3f3f) ans = -1;
        cout << ans << endl;
    }
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

题意:给定一个字符串要去判断变成回文串的最小次数。要求只准删除一种字母

通过研究小的样例我们发现每次可以比较首尾两个，如果不同，则删除不同的预定的字母，否则都删除知道最后大小<=1