BZOJ 1396: 识别子串( 后缀数组 + 线段树 )

这道题各位大神好像都是用后缀自动机做的?.....蒟蒻就秀秀智商写一写后缀数组解法.....

求出Height数组后, 我们枚举每一位当做子串的开头.



如上图(x, y是height值), Heights数组中相邻的3个后缀, 假如我们枚举s2的第一个字符为开头, 那我们发现, 长度至少为len = max(x, y)+1, 才能满足题意(仅出现一次). 这个很好脑补...因为s2和其他串的LCP是RMQ, 肯定会<=LCP(s1,s2)或<=LCP(s2,s3). 然后就用len去更新s2中前len个字符的答案, 线段树维护. 然后对于长度lth>len的也肯定是合法的, 他们对s2的前lth个字符都有贡献...但是事实上lth对前lth-1个字符c的贡献是没有卵用的....(因为小于同样字符开头的以c结尾的串的贡献或者是len的贡献), 所以lth>=len对第lth个字符有贡献.



容易看出这样的贡献是成等差数列的。。。。线段树维护就OK了.

时间复杂度O(N log N), 空间复杂度O(N)

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#include<cstdio>#include<cstring>#include<algorithm> using namespace std; const int maxn = 100009; char S[maxn];int N, L, R, Val;int Height[maxn], Rank[maxn], Sa[maxn], cnt[maxn]; inline void Min(int &x, int t) { if(t < x) x = t;}inline void Max(int &x, int t) { if(t > x) x = t;} void BuildSA(int m) { int *x = Height, *y = Rank; for(int i = 0; i < m; i++) cnt[i] = 0; for(int i = 0; i < N; i++) cnt[x[i] = S[i]]++; for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1]; for(int i = N; i--; ) Sa[--cnt[x[i]]] = i; for(int k = 1, p = 0; k <= N; k <<= 1, p = 0) { for(int i = N - k; i < N; i++) y[p++] = i; for(int i = 0; i < N; i++) if(Sa[i] >= k) y[p++] = Sa[i] - k; for(int i = 0; i < m; i++) cnt[i] = 0; for(int i = 0; i < N; i++) cnt[x[y[i]]]++; for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1]; for(int i = N; i--; ) Sa[--cnt[x[y[i]]]] = y[i]; swap(x, y); p = (x[Sa[0]] = 0) + 1; for(int i = 1; i < N; i++) { if(y[Sa[i]] != y[Sa[i - 1]] || y[Sa[i] + k] != y[Sa[i - 1] + k]) p++; x[Sa[i]] = p - 1; } if((m = p) >= N) break; } for(int i = 0; i < N; i++) Rank[Sa[i]] = i; Height[0] = 0; for(int i = 0, h = 0; i < N; i++) if(Rank[i]) { if(h) h--; while(S[i + h] == S[Sa[Rank[i] - 1] + h]) h++; Height[Rank[i]] = h; }} struct Node { Node *lc, *rc; int n, d; inline void pd(int len) { if(n != maxn) { Min(lc->n, n); Min(rc->n, n); } if(d != maxn) { Min(lc->d, d); Min(rc->d, d + ((len + 1) >> 1)); } }} pool[maxn << 1], *pt = pool, *Root; void Build(Node* t, int l, int r) { t->n = t->d = maxn; if(l != r) { int m = (l + r) >> 1; Build(t->lc = pt++, l, m); Build(t->rc = pt++, m + 1, r); }} void Modify(Node* t, int l, int r) { if(L <= l && r <= R) { Min(t->n, Val); } else { int m = (l + r) >> 1; if(L <= m) Modify(t->lc, l, m); if(m < R) Modify(t->rc, m + 1, r); }} void Change(Node* t, int l, int r) { if(L <= l && r <= R) { Min(t->d, Val + l - L); } else { int m = (l + r) >> 1; if(L <= m) Change(t->lc, l, m); if(m < R) Change(t->rc, m + 1, r); }} void DFS(Node* t, int l, int r) { if(l != r) { int m = (l + r) >> 1; t->pd(r - l + 1); DFS(t->lc, l, m); DFS(t->rc, m + 1, r); } else printf("%d\n", min(t->d, t->n));} int main() { scanf("%s", S); N = strlen(S); S[N++] = '$'; BuildSA('z' + 1); int n = N - 1; Build(Root = pt++, 1, n); Height
= 0; for(int i = 1; i < N; i++) { Val = max(Height[i], Height[i + 1]) + 1; if(Val > 1) { if(Sa[i] + Val > n) continue; L = Sa[i] + 1, R = L + Val - 2; Modify(Root, 1, n); } L = Sa[i] + Val, R = n; if(L > R) continue; Change(Root, 1, n); } DFS(Root, 1, n); return 0;}-------------------------------------------------------------------------

1396: 识别子串

Time Limit: 10 Sec Memory Limit: 162 MB
Submit: 201 Solved: 119
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Description

Input

一行,一个由小写字母组成的字符串S,长度不超过10^5

Output

L行,每行一个整数,第i行的数据表示关于S的第i个元素的最短识别子串有多长.

Sample Input

agoodcookcooksgoodfood

Sample Output

1
2
3
3
2
2
3
3
2
2
3
3
2
1
2
3
3
2
1
2
3
4

HINT

Source