这题又是万恶的线段树 maxx[j]存储的是 l = xxx, r = j的时候的答案 我们会让 l 从 1到n 的遍历中，查询线段树的[l, n]中最大的答案 因为query的下界是n，所以单次查询复杂度是logn 再其次这样做必须得再每次单元操作之后 对线段树 进行update
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define ll long long #define ld long double #define null NULL #define all(a) a.begin(), a.end() #define forn(i, n) for (int i = 0; i < n; ++i) #define sz(a) (int)a.size() #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 template int gmax(T &a, T b) { if (b > a) { a = b; return 1; } return 0; } template int gmin(T &a, T b) { if (b < a) { a = b; return 1; } return 0; } using namespace std; int n, s; const int N = 1e5 + 5; int A[N]; // origin input array vector pos[N]; // positions in initial array split by its value int nowSize[N]; // the order of A[i] in their own position array int input[N]; // use for segment tree input int maxx[N << 2]; int lazy[N << 2]; void pushUp(int rt) { maxx[rt] = max(maxx[rt << 1], maxx[rt << 1 | 1]); } void pushDown(int rt) { if(lazy[rt]) { maxx[rt << 1] += lazy[rt]; maxx[rt << 1 | 1] += lazy[rt]; lazy[rt << 1] += lazy[rt]; lazy[rt << 1 | 1] += lazy[rt]; lazy[rt] = 0; } } void build(int l, int r, int rt) { lazy[rt] = 0; if(l == r) { maxx[rt] = input[l]; return; } int m = (l + r) >> 1; build(lson); build(rson); pushUp(rt); } int query(int L, int R, int l, int r, int rt) { if(L <= l && r <= R) return maxx[rt]; pushDown(rt); int m = (l + r) >> 1; int ret = -1; if(L <= m) ret = max(ret, query(L, R, lson)); if(R > m) ret = max(ret, query(L, R, rson)); return ret; } void update(int c, int L, int R, int l, int r, int rt) { if(L <= l && r <= R) { maxx[rt] += c; lazy[rt] += c; return; } pushDown(rt); int m = (l + r) >> 1; if(L <= m) update(c, L, R, lson); if(R > m) update(c, L, R, rson); pushUp(rt); } int main() { int T; scanf("%d", &T); for(int cas = 1; cas <= T; ++ cas) { scanf("%d %d", &n, &s); for(int i = 0; i < N; ++i) pos[i].clear(); for(int i = 1; i <= n; ++i) scanf("%d", &A[i]); for(int i = 1; i <= n; ++i) { pos[A[i]].push_back(i); nowSize[i] = pos[A[i]].size(); } for(int i = 0; i < N; ++i) pos[i].push_back(n + 1); input[0] = 0; for(int i = 1; i <= n; ++i) { input[i] = input[i-1] + 1; if(nowSize[i] == s + 1) input[i] -= s + 1; else if(nowSize[i] > s + 1) input[i] --; } build(1, n, 1); int ans = -1; for(int i = 1; i <= n; ++i) { ans = max(ans, query(i, n, 1, n, 1)); int num = A[i]; if(nowSize[i] + s < pos[num].size()) { int tmp = pos[num][nowSize[i] + s - 1]; int tmp2 = pos[num][nowSize[i] + s]; update(-1, i, tmp - 1, 1, n, 1); update(s, tmp, tmp2 - 1, 1, n, 1); } else { update(-1, i, n, 1, n, 1); } } printf("Case #%d: %d\n", cas, ans); } return 0; }
