4556: [Tjoi2016&Heoi2016]字符串
分析:
首先可以二分这个长度。此时需要判断是否存在一个以b结尾的前缀,满足与[c,d]的lcp大于等于mid。
如果我们把串翻转,那么就是判断是否存在一个以b开始的后缀,这样可以建出SAM,线段树维护每个点的right集合。此时在从包含[1,b]这个状态的点沿着parent树往上跳,那么长度会减少,相应的right集合会变大,于是可以跳到第一个长度满足的点,right集合也是满足的点中最大的,看这个点中是否存在[a,b-mid+1]的位置。
SA的做法:求出height数组,二分一个长度mid,找到[c,d]的rnk设为p,从p往左往右找满足lcp长度都大于等于mid的区间[l,r],然后判断[l,r]中是否在[a,b-mid+1]中出现过。
代码:
#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> #include<cmath> #include<cctype> #include<set> #include<queue> #include<vector> #include<map> using namespace std; typedef long long LL; inline int read() { int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1; for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f; } const int N = 200005; int ls[N * 20], rs[N * 20], Root[N], pos[N], ch[N][26], len[N], fa[N], tmp[N], xl[N], f[N][20]; int TreeIndex, Index = 1, Last = 1, n; char s[N]; void Insert(int l,int r,int &now,int p) { now = ++TreeIndex; if (l == r) return ; int mid = (l + r) >> 1; if (p <= mid) Insert(l, mid, ls[now], p); else Insert(mid + 1, r, rs[now], p); } int query(int l,int r,int now,int L,int R) { if (!now) return 0; if (L <= l && r <= R) return 1; int mid = (l + r) >> 1; if (L <= mid && query(l, mid, ls[now], L, R)) return 1; if (R > mid && query(mid + 1, r, rs[now], L, R)) return 1; return 0; } int Merge(int x,int y) { if (!x || !y) return x + y; int z = ++TreeIndex; ls[z] = Merge(ls[x], ls[y]); rs[z] = Merge(rs[x], rs[y]); return z; } void extend(int c,int i) { int np = ++Index, p = Last; pos[i] = np; // pos[i] = np; !!!!! len[np] = len[p] + 1; for (; p && !ch[p][c]; p = fa[p]) ch[p][c] = np; if (!p) fa[np] = 1; else { int q = ch[p][c]; if (len[q] == len[p] + 1) fa[np] = q; else { int nq = ++Index; memcpy(ch[nq], ch[q], sizeof(ch[q])); fa[nq] = fa[q]; fa[np] = fa[q] = nq; len[nq] = len[p] + 1; for (; p && ch[p][c] == q; p = fa[p]) ch[p][c] = nq; } } Last = np; Insert(1, n, Root[np], i); } bool check(int mid,int x,int l,int r) { for (int i = 19; ~i; --i) if (len[f[x][i]] >= mid) x = f[x][i]; return query(1, n, Root[x], l, r); } int main() { n = read();int m = read(); scanf("%s", s + 1); reverse(s + 1, s + n + 1); for (int i = 1; i <= n; ++i) extend(s[i] - 'a', i); for (int i = 1; i <= Index; ++i) tmp[len[i]] ++; for (int i = 1; i <= n; ++i) tmp[i] += tmp[i - 1]; for (int i = Index; i >= 1; --i) xl[tmp[len[i]]--] = i; for (int i = Index; i > 1; --i) { int x = xl[i]; Root[fa[x]] = Merge(Root[fa[x]], Root[x]); } for (int i = 1; i <= Index; ++i) f[i][0] = fa[i]; for (int j = 1; j <= 19; ++j) for (int i = 1; i <= Index; ++i) f[i][j] = f[f[i][j - 1]][j - 1]; while (m --) { int a = n - read() + 1, b = n - read() + 1, c = n - read() + 1, d = n - read() + 1; swap(a, b); swap(c, d); int l = 1, r = min(d - c + 1, b - a + 1), ans = 0; while (l <= r) { int mid = (l + r) >> 1; if (check(mid, pos[d], a + mid - 1, b)) ans = mid, l = mid + 1; else r = mid - 1; } printf("%d ", ans); } return 0; }