[POI2000]公共串 - 后缀数组

Update 2018.1.8：sto lsy orz 给出了一个 \(O(n)\) 的做法。

Description

求若干个串的最长的公共子串的长度。

Solution

考虑将这若干个串全部拼起来，中间用一些不在字符集内的符号隔开。

然后二分答案 \(K\)，如果连续的一段 \(height\) 都大于等于 \(K\)，且每个串都出现了至少一次，则是可行的。

O(N)

并没有必要二分答案，前面是二分长度，然后 \(check\) 一下是否有一段连续的 \(height\) 区间使得每个值都大于等于 \(K\)，且每个串都出现至少一次。

反过来，我们可以枚举左端点，然后不断向右扩展右端点，当每种串都出现恰好至少一次时，就对把答案和这一段区间的 \(\min\) 取个 \(\max\)，然后区间的 \(\min\) 用单调队列维护即可。

然后可能是由于我的代码比较丑陋，且在洛谷上此题数据范围比较小，我的 \(O(N)\) 居然跑得比 \(O(N\log N)\) 慢。（笑

Code

O(NlogN)​

#include <bits/stdc++.h>
using namespace std;

const int _ = 1e5 + 10;
int N, n, s[_], belong[_];
int rnk[_], sa[_], height[_];
char str[_];

void SA() {
	static int t[_], a[_], buc[_], fir[_], sec[_], tmp[_];
	copy(s + 1, s + N + 1, t + 1);
	sort(t + 1, t + N + 1);
	int *end = unique(t + 1, t + N + 1);
	for (int i = 1; i <= N; ++i) a[i] = lower_bound(t + 1, end, s[i]) - t;
	fill(buc + 1, buc + N + 1, 0);
	for (int i = 1; i <= N; ++i) ++buc[a[i]];
	for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
	for (int i = 1; i <= N; ++i) rnk[i] = buc[a[i] - 1] + 1;
	for (int len = 1; len <= N; len <<= 1) {
		for (int i = 1; i <= N; ++i) {
			fir[i] = rnk[i];
			sec[i] = i + len > N ? 0 : rnk[i + len];
		}
		fill(buc + 1, buc + N + 1, 0);
		for (int i = 1; i <= N; ++i) ++buc[sec[i]];
		for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
		for (int i = 1; i <= N; ++i) tmp[N - --buc[sec[i]]] = i;
		fill(buc + 1, buc + N + 1, 0);
		for (int i = 1; i <= N; ++i) ++buc[fir[i]];
		for (int i = 1; i <= N; ++i) buc[i] += buc[i - 1];
		for (int i, j = 1; j <= N; ++j) {
			i = tmp[j];
			sa[buc[fir[i]]--] = i;
		}
		bool same = false;
		for (int i, j = 1, last = 0; j <= N; ++j) {
			i = sa[j];
			if (!last) rnk[i] = 1;
			else if (fir[i] == fir[last] && sec[i] == sec[last])
				rnk[i] = rnk[last], same = true;
			else rnk[i] = rnk[last] + 1;
			last = i;
		}
		if (!same) break;
	}
	for (int i = 1, k = 0; i <= N; ++i) {
		if (rnk[i] == 1) k = 0;
		else {
			if (k > 0) --k;
			int j = sa[rnk[i] - 1];
			while (i + k <= N && j + k <= N && a[i + k] == a[j + k]) ++k;
		}
		height[rnk[i]] = k;
	}
}


bool check(int k) {
	static int vis[_], tot = 0;
	int cnt = 0;
	++tot;
	for (int i = 1; i <= N; ++i) {
		if (height[i] < k) cnt = 0, ++tot;
		else {
			if (vis[belong[sa[i]]] != tot)
				vis[belong[sa[i]]] = tot, ++cnt;
			if (vis[belong[sa[i - 1]]] != tot)
				vis[belong[sa[i - 1]]] = tot, ++cnt;
			if (cnt == n) return true;
		}
	}
	return false;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("string.in", "r", stdin);
	freopen("string.out", "w", stdout);
#endif
	scanf("%d", &n);
	int now = 0;
	for (int i = 1; i <= n; ++i) {
		++now;
		scanf("%s", str);
		int len = strlen(str);
		for (int j = now; j <= now + len - 1; ++j)
			s[j] = str[j - now] - 'a' + 1, belong[j] = i;
		now += len - 1;
		s[++now] = i + 26;
	}
	N = now;
	SA();
	int l = 0, r = N;
	while (l < r) {
		int mid = (l + r + 1) >> 1;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}
	printf("%d\n", l);
	return 0;
}

O(N)

#include <bits/stdc++.h>
using namespace std;

const int _ = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int N, n, s[_], belong[_];
int rk[_], sa[_], height[_];
char str[_];

inline void SA() {
  int M = 125, p = 0;  // 字符集
  static int buc[_], id[_], fir[_], t[_], oldrk[_];
  copy(s + 1, s + N + 1, t + 1);
  sort(t + 1, t + N + 1);
  int *end = unique(t + 1, t + N + 1);
  for (int i = 1; i <= N; ++i) s[i] = lower_bound(t + 1, end, s[i]) - t;
  for (int i = 1; i <= N; ++i) ++buc[rk[i] = s[i]];
  for (int i = 1; i <= M; ++i) buc[i] += buc[i - 1];
  for (int i = N; i >= 1; --i) sa[buc[rk[i]]--] = i;
  for (int w = 1; w < N; w <<= 1, M = p) {
    p = 0;
    for (int i = N; i > N - w; --i) id[++p] = i;
    for (int i = 1; i <= N; ++i)
      if (sa[i] > w) id[++p] = sa[i] - w;
    fill(buc + 1, buc + M + 1, 0);
    for (int i = 1; i <= N; ++i) ++buc[fir[i] = rk[id[i]]];
    for (int i = 1; i <= M; ++i) buc[i] += buc[i - 1];
    for (int i = N; i >= 1; --i) sa[buc[fir[i]]--] = id[i];
    copy(rk + 1, rk + N + 1, oldrk + 1);
    p = 0;
    for (int i = 1; i <= N; ++i) {
      int x = sa[i], y = sa[i - 1];
      rk[sa[i]] =
          (oldrk[x] == oldrk[y] && oldrk[x + w] == oldrk[y + w]) ? p : ++p;
    }
    if (p == N) break;
  }
  for (int i = 1, k = 0; i <= N; ++i) {
    if (rk[i] == 1)
      k = 0;
    else {
      if (k > 0) --k;
      int j = sa[rk[i] - 1];
      while (i + k <= N && j + k <= N && s[i + k] == s[j + k]) ++k;
    }
    height[rk[i]] = k;
  }
}

int ans = 0;
void solve() {
  static int vis[_], cnt = 0;
  static int q[_];
  int l = 1, r = 1;
  for (int i = 1, j = 1; i <= N; ++i) {
    while (l < r && q[l] < i) ++l;
    while (j <= N && cnt < n) {
      if (!vis[belong[sa[j]]]) ++cnt;
      ++vis[belong[sa[j]]];
      while (l < r && height[q[r - 1]] >= height[j]) --r;
      q[r++] = j;
      ++j;
    }
    if (cnt == n)
      ans = max(ans, height[q[l]]);
    else
      return;
    if (i - 1 > 0) {
      --vis[belong[sa[i - 1]]];
      if (!vis[belong[sa[i - 1]]]) --cnt;
    }
  }
}

int main() {
#ifndef ONLINE_JUDGE
  freopen("string.in", "r", stdin);
  freopen("string.out", "w", stdout);
#endif
  scanf("%d", &n);
  int now = 0;
  for (int i = 1; i <= n; ++i) {
    ++now;
    scanf("%s", str);
    int len = strlen(str);
    for (int j = now; j <= now + len - 1; ++j)
      s[j] = str[j - now] - 'a' + 1, belong[j] = i;
    now += len - 1;
    s[++now] = i + 26;
  }
  N = now;
  SA();
  solve();
  printf("%d\n", ans);
  return 0;
}