标签:sync uil 线段 pos lin san pll 强制 cond
因为强制在线所以只能转成序列上的问题然后树套树了。。。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned 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 ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 2e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} int r[N], sa[N], _t[N], _t2[N], c[N], rk[N], lcp[N], san; void buildSa(int *r, 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] = r[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(r[i + k] == r[j + k]) k++; lcp[rk[i]] = k; } } struct bit_seg { int Rt[N], ls[N * 50], rs[N * 50], sum[N * 50], tot, n, maxval; void init(int _n, int _maxval) { n = _n; maxval = _maxval; for(int i = 0; i <= tot; i++) sum[i] = ls[i] = rs[i] = 0; for(int i = 0; i <= n; i++) Rt[i] = 0; tot = 0; } void update(int& o, int pos, int v, int l, int r) { if(!o) o = ++tot; sum[o] += v; if(l == r)return ; int m = l + r >> 1; if(pos <= m) update(ls[o], pos, v, l, m); else update(rs[o], pos, v, m + 1, r); } int query(int o, int L, int R, int l, int r) { if(L > R || l > r || !o) return 0; if(R < l || r < L) return 0; if(L <= l && r <= R) return sum[o]; int m = l + r >> 1; return query(ls[o] , L , R , l , m) + query(rs[o], L, R, m + 1, r); } int bitquery(int L, int R, int a, int b) { int ans = 0; for(int i = R; i; i -= i & -i) ans += query(Rt[i], a, b, 0, maxval); for(int i = L - 1; i; i -= i & -i) ans -= query(Rt[i], a, b, 0, maxval); return ans; } void bitupdate(int i,int pos,int v) { for( ; i <= n; i += i & -i) update(Rt[i], pos, v, 0, maxval); } } S; int Log[N]; struct ST { int dp[N][20], ty; void build(int n, int b[], int _ty) { ty = _ty; 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]); } } rmq; bool is[N]; char s[N], t[N]; int f[N], q; int main() { for(int i = -(Log[0]=-1); i < N; i++) Log[i] = Log[i - 1] + ((i & (i - 1)) == 0); int T; scanf("%d", &T); while(T--) { san = 0; scanf("%s%s", s, t); int lens = strlen(s), lent = strlen(t); reverse(s, s + lens); reverse(t, t + lent); for(int i = 0; i < lens; i++) scanf("%d", &f[i]); reverse(f, f + lens); for(int i = 0; i < lens; i++) { is[san] = true; r[san++] = s[i]; } is[san] = false; r[san++] = ‘$‘; for(int i = 0; i < lent; i++) { is[san] = false; r[san++] = t[i]; } is[san] = false; r[san] = 0; S.init(san, lens - 1); buildSa(r, san + 1, 256); rmq.build(san, lcp, -1); for(int i = 1; i <= san; i++) if(is[sa[i]]) S.bitupdate(i, sa[i], f[sa[i]]); scanf("%d", &q); int a = 0, b = 0, c = 0, d = 0, add = 0, ans = 0, len = 0; int low = 0, high = 0, mid, L = 0, R = 0; while(q--) { int op; scanf("%d", &op); if(op == 1) { scanf("%d%d", &a, &b); a ^= ans; b ^= ans; a = lens - a - 1; if(a >= lens) { return 0; } add = b - f[a]; f[a] = b; S.bitupdate(rk[a], a, add); } else { scanf("%d%d%d%d", &c, &d, &a, &b); a ^= ans; b ^= ans; c ^= ans; d ^= ans; a = lens - a - 1; b = lens - b - 1; c = lent - c - 1 + lens + 1; d = lent - d - 1 + lens + 1; swap(a, b); swap(c, d); len = d - c + 1; low = 1, high = rk[c] - 1, L = rk[c], R = rk[c]; while(low <= high) { mid = low + high >> 1; if(rmq.query(mid + 1, rk[c]) >= len) high = mid - 1, L = mid; else low = mid + 1; } low = rk[c] + 1, high = san; while(low <= high) { mid = low + high >> 1; if(rmq.query(rk[c] + 1, mid) >= len) low = mid + 1, R = mid; else high = mid - 1; } b = b - len + 1; if(a > b) ans = 0; else ans = S.bitquery(L, R, a, b); printf("%d\n", ans); } } } return 0; } /* */
标签:sync uil 线段 pos lin san pll 强制 cond
原文地址:https://www.cnblogs.com/CJLHY/p/10767297.html