思路
- 首先是要选出 L a r g e s t S u b s e q u e n c e Largest Subsequence LargestSubsequence ,第一个字符是整个串最大的,第二个是第一个的位置后面最大的 … \dots … ,我用的是优先队列第一关键字大小,第二关键字位置。(其实不用这样,用栈就行,我复杂了)
- 然后你会发现其 c y c l i c cyclic cyclic 次数是字符串长度减 1 1 1 ,过程就是交换首位字符的位置,向中心扩展。
- 有可能你选出来的字符串从第一个开始有连续一样的,如: z z z d c zzzdc zzzdc , z z z zzz zzz 就是连续一样的,试想如果是这样还是需要字符串长度减 1 1 1 次?当然不是,而是长度减去连续的长度。
- 最后结束检查一遍是否 s o r t e d sorted sorted
Think Twice, Code Once
#include <bits/stdc++.h>
#define il inline
#define get getchar
#define put putchar
#define is isdigit
#define int long long
#define dfor(i,a,b) for(int i=a;i<=b;++i)
#define dforr(i,a,b) for(int i=a;i>=b;--i)
#define dforn(i,a,b) for(int i=a;i<=b;++i,put(10))
#define mem(a,b) memset(a,b,sizeof a)
#define memc(a,b) memcpy(a,b,sizeof a)
#define pr 114514191981
#define gg(a) cout<<a,put(32)
#define INF 0x7fffffff
#define tt(x) cout<<x<<'\n'
#define endl '\n'
#define ls i<<1
#define rs i<<1|1
#define la(r) tr[r].ch[0]
#define ra(r) tr[r].ch[1]
#define lowbit(x) (x&-x)
#define ct cin.tie(nullptr),ios_base::sync_with_stdio(false)
using namespace std;
typedef unsigned int ull;
typedef pair<int, int> pii;
int read(void) {int x=0,f=1;char c=get();while(!is(c)) (f=c==45?-1:1),c=get();while(is(c)) x=(x<<1)+(x<<3)+(c^48),c=get();return x*f;
}
void write(int x) {if (x < 0) x = -x, put(45);if (x > 9) write(x / 10);put((x % 10) ^ 48);
}
#define writeln(a) write(a), put(10)
#define writesp(a) write(a), put(32)
#define writessp(a) put(32), write(a)
const int N = 1e5 + 10, M = 1e5 + 10, SN = 1e3 + 10, mod = 998244353;
struct p {char s;int pos;il bool operator <(const p & b) const {return s < b.s || (s == b.s && pos > b.pos);}
};
signed main() {int T = read();while (T--) {int n = read();priority_queue<p> q;string s;cin >> s;bool f2 = 1;for (int i = 0; i < s.size(); ++i) q.push({s[i], i});vector<p> a;while (!q.empty()) {a.push_back(q.top());int pos = q.top().pos; q.pop();while (!q.empty() && q.top().pos < pos) q.pop();}int l = 0, r = a.size() - 1, cnt = a.size() >> 1;int cnt1 = 0;for (int i = 0; i < a.size(); ++i) {if (a[i].s == a[0].s) ++cnt1;}while (cnt--) swap(s[a[l++].pos], s[a[r--].pos]);for (int i = 1; i < s.size(); ++i) {if (s[i] < s[i - 1]) {f2 = 0; break;}}f2? (write(a.size() - cnt1), put(10)): puts("-1");}return 0;
}