Codeforces 653F Paper task SA
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如果不要求本质不同直接st表二分找出最右端, 然后计数就好了。
要求本质不同, 先求个sa, 然后用lcp求本质不同就好啦。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 5e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-6; const double PI = acos(-1); int Log[N]; struct ST { int dp[N][20], ty; void build(int n, int b[], int _ty) { ty = _ty; for(int i = -(Log[0]=-1); i < N; i++) Log[i] = Log[i - 1] + ((i & (i - 1)) == 0); for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i]; for(int j = 1; j <= Log[n]; j++) for(int i = 1; i + (1 << j) - 1 <= n; i++) dp[i][j] = max(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]); } int query(int x, int y) { int k = Log[y - x + 1]; return ty * max(dp[x][k], dp[y - (1 << k) + 1][k]); } }; const int MAX = 6e5; const int MIN = -6e5; namespace SGT { int tot = 0, Rt[N]; struct Node { int sum, ls, rs; } a[N * 25]; void modify(int p, int l, int r, int& x, int y) { x = ++tot; a[x] = a[y]; a[x].sum++; if(l == r) return; int mid = l + r >> 1; if(p <= mid) modify(p, l, mid, a[x].ls, a[y].ls); else modify(p, mid + 1, r, a[x].rs, a[y].rs); } int query(int p, int l, int r, int x) { if(l == r) return a[x].sum; int mid = l + r >> 1; if(p <= mid) query(p, l, mid, a[x].ls); else query(p, mid + 1, r, a[x].rs); } } int sa[N], t[N], t2[N], c[N], rk[N], lcp[N]; void buildSa(char *s, int n, int m) { int i, j = 0, k = 0, *x = t, *y = t2; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[i] = s[i]]++; for(i = 1; i < m; i++) c[i] += c[i - 1]; for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1) { int p = 0; for(i = n - k; i < n; i++) y[p++] = i; for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k; for(i = 0; i < m; i++) c[i] = 0; for(i = 0; i < n; i++) c[x[y[i]]]++; for(i = 1; i < m; i++) c[i] += c[i - 1]; for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(int i = 1; i < n; i++) { if(y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) x[sa[i]] = p - 1; else x[sa[i]] = p++; } if(p >= n) break; m = p; } for(i = 1; i < n; i++) rk[sa[i]] = i; for(i = 0; i < n - 1; i++) { if(k) k--; j = sa[rk[i] - 1]; while(s[i + k] == s[j + k]) k++; lcp[rk[i]] = k; } } int n, a[N], p[N]; char s[N]; ST rmq; int main() { scanf("%d", &n); scanf("%s", s + 1); for(int i = 1; i <= n; i++) { a[i] = s[i] == ‘(‘ ? 1 : -1; a[i] += a[i - 1]; } rmq.build(n, a, -1); for(int i = 1; i <= n; i++) SGT::modify(a[i], MIN, MAX, SGT::Rt[i], SGT::Rt[i - 1]); for(int i = 1; i <= n; i++) { if(s[i] == ‘)‘) continue; int low = i + 1, high = n; p[i] = i; while(low <= high) { int mid = low + high >> 1; if(rmq.query(i, mid) >= a[i] - 1) p[i] = mid, low = mid + 1; else high = mid - 1; } } LL ans = 0; buildSa(s + 1, n + 1, 255); for(int i = 1; i <= n; i++) { int x = sa[i] + 1; if(s[x] == ‘)‘) continue; int dn = x + lcp[i], up = p[x]; if(dn <= up) { ans += SGT::query(a[x] - 1, MIN, MAX, SGT::Rt[up]); ans -= SGT::query(a[x] - 1, MIN, MAX, SGT::Rt[dn - 1]); } } printf("%lld\n", ans); return 0; } /* */
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