CodeForces 1706C Qpwoeirut And The City
Qpwoeirut And The City 对于奇数长度的数组，为了让合法的元素最多，只有一种情况。 对于偶数长度的数组，如果把合法的点标记为 $1$，其它点标记为 $0$，可以发现 $0$ 和 $1$ 的排列规律，首先不能出现连续的 $1$，并且最多出现一次连续的两个 $0$，连续的两个 $0$ 出现的位置决定了最终的结果。 从 $a_2$ 开始，假设 $a_2 = 1, a_3 = 0, \ldots$ 这是一种情况。从 $a_{n - 1}$ 开始，假设 $a_{n - 1} = 1, a_{n - 2} = 0, \ldots$ 这是对称的另外一种情况。对这两种情况分别求出前缀和与后缀和，然后枚举 $i$，就可以求出答案。 // 2024/1/29 #include <climits> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <functional> #include <iomanip> #include <iostream> #include <map> #include <queue> #include <set> #include <sstream> #include <stack> #include <string> #include <utility> #include <vector> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef vector<int> VI; typedef pair<int, int> PII; template <class T> using pq = priority_queue<T>; template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>; const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353; const ll INFL = 0x3f3f3f3f'3f3f3f3f; const double eps = 1e-8; const int dir[8][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}, }; const ull Pr = 131; #define For(i, a, b) for (int i = int(a); i < int(b); ++i) #define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i) #define For1(i, a, b) for (int i = int(a); i <= int(b); ++i) #define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i) #define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i]) #define f1 first #define f2 second #define pb push_back #define has(a, x) (a.find(x) != a.end()) #define nemp(a) (!a.empty()) #define all(a) (a).begin(), (a).end() #define SZ(a) int((a).size()) #define NL cout << '\n'; template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; } template <typename t> istream &operator>>(istream &in, vector<t> &vec) { for (t &x : vec) in >> x; return in; } template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) { int n = SZ(vec); For(i, 0, n) { out << vec[i]; if (i < n - 1) out << ' '; } return out; } void __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\'' << x << '\''; } void __print(const char *x) { cerr << '\"' << x << '\"'; } void __print(const string &x) { cerr << '\"' << x << '\"'; } void __print(bool x) { cerr << (x ? "true" : "false"); } template <typename T, typename V> void __print(const pair<T, V> &x) { cerr << '{'; __print(x.first); cerr << ", "; __print(x.second); cerr << '}'; } template <typename T> void __print(const T &x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? ", " : ""), __print(i); cerr << "}"; } void _print() { cerr << "]\n"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef _DEBUG #define debug1(x) cout << #x " = " << x << endl; #define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl; #define debug3(x, y, z) \ cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl; #define dbg(x...) \ cerr << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << endl; #else #define debug1 #define debug2 #define debug3 #define dbg(x...) #endif const int N = 300100; int n; ll a[N], p[N], p1[N]; void solve() { cin >> n; For1(i, 1, n) { cin >> a[i]; } ll ans{}; auto check = [](int start) { ll ans = 0; for (int i = start; i <= n - 1; i += 2) { if (a[i] > a[i - 1] && a[i] > a[i + 1]) continue; ll top = max(a[i - 1], a[i + 1]) + 1; ans += top - a[i]; } return ans; }; if (n & 1) { ans = check(2); cout << ans << '\n'; return; } p[1] = 0; for (int i = 2; i <= n; i += 2) { ll top = max(a[i - 1], a[i + 1]) + 1, delta = max(top - a[i], 0LL); p[i] = p[i + 1] = delta + p[i - 1]; } p1[n] = 0; for (int i = n - 1; i >= 1; i -= 2) { ll top = max(a[i - 1], a[i + 1]) + 1, delta = max(top - a[i], 0LL); p1[i] = p1[i - 1] = delta + p1[i + 1]; } ans = -1; for (int i = 2; i <= n; i += 2) { ll tmp = p[i - 1] + p1[i]; if (ans == -1) ans = tmp; else ckmin(ans, tmp); } cout << ans << '\n'; } int main(void) { #ifdef _DEBUG // freopen("1706c.in", "r", stdin); #endif std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int T = 1; cin >> T; while (T--) { solve(); } return 0; }