A 重新排列
题目地址:
基本思路:
尺取法,然后每次一下范围内的字母数量能否构成”puleyaknoi“就行了。
参考代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #include <bits/stdc++.h> using namespace std; #define IO std::ios::sync_with_stdio(false); cin.tie(0) #define int long long #define ull unsigned long long #define SZ(x) ((int)(x).size()) #define all(x) (x).begin(), (x).end() #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 mset(s, _) memset(s, _, sizeof(s)) #define pb push_back #define pii pair <int, int> #define mp(a, b) make_pair(a, b) #define INF 0x3f3f3f3f inline int read() { int x = 0, neg = 1; char op = getchar(); while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); } while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); } return neg * x; } inline void print(int x) { if (x < 0) { putchar('-'); x = -x; } if (x >= 10) print(x / 10); putchar(x % 10 + '0'); } string mark = "puleyaknoi"; int memo[26],now[26]; string str; bool check(){ for(int i = 0; i < 26 ; i++) if(now[i] < memo[i]) return false; return true; } signed main() { IO; for (auto s : mark) memo[s - 'a']++; int t; cin >> t; while (t--){ cin >> str; int n = SZ(str); str = ' ' + str; int r = 1,ans = INF; now[str[r] - 'a']++; for(int l = 1 ; l <= n ; l++) { while (!check() && r + 1 <= n) { now[str[++r] - 'a']++; } if (check()) ans = min(ans, r - l + 1); now[str[l] - 'a']--; } if(ans == INF) cout << -1 << '\n'; else cout << ans << '\n'; } return 0; }