題目鏈接

思路

求不可重疊最長重復子串，也可以利用$height[i]$。

• 二分枚舉答案$K$
• 當$height[i]\ge K$可能存在解，然后判斷$min(sa[i],sa[i?1])，max(sa[i],sa[i?1])$的關系，看能否構造成不重疊的公共前綴
• $height[i]$數值相同的分成一組，取小組中的最值計算

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define mid ((l + r) >> 1) #define lc rt<<1 #define rc rt<<1|1 #define endl '\n' const int maxn = 1e5 + 5; const int inf = 0x3f3f3f3f; const int mod = 1e9 + 7; using namespace std; int cntA[maxn], cntB[maxn], A[maxn], B[maxn]; int Sa[maxn], tsa[maxn], height[maxn], Rank[maxn]; int s[maxn]; int n; void suffixArray () {for (int i = 0; i < 1000; ++i) cntA[i] = 0;for (int i = 1; i <= n; ++i) cntA[s[i]]++;for (int i = 1; i < 1000; ++i) cntA[i] += cntA[i-1];for (int i = n; i >= 1; --i) Sa[ cntA[s[i]]-- ] = i;Rank[ Sa[1] ] = 1;for (int i = 2; i <= n; ++i) {Rank[Sa[i]] = Rank[Sa[i-1]];if (s[Sa[i]] != s[Sa[i-1]]) Rank[Sa[i]]++;}for (int l = 1; Rank[Sa[n]] < n; l <<= 1) {for (int i = 0; i <= n; ++i) cntA[i] = 0;for (int i = 0; i <= n; ++i) cntB[i] = 0;for (int i = 1; i <= n; ++i) {cntA[ A[i] = Rank[i] ]++;cntB[ B[i] = (i + l <= n) ? Rank[i+l] : 0]++;}for (int i = 1; i <= n; ++i) cntB[i] += cntB[i-1];for (int i = n; i >= 1; --i) tsa[ cntB[B[i]]-- ] = i;for (int i = 1; i <= n; ++i) cntA[i] += cntA[i-1];for (int i = n; i >= 1; --i) Sa[ cntA[A[tsa[i]]]-- ] = tsa[i];Rank[ Sa[1] ] = 1;for (int i = 2; i <= n; ++i) {Rank[Sa[i]] = Rank[Sa[i-1]];if (A[Sa[i]] != A[Sa[i-1]] || B[Sa[i]] != B[Sa[i-1]]) Rank[Sa[i]]++;}}for (int i = 1, j = 0; i <= n; ++i) {if (j) --j;int tmp = Sa[Rank[i] - 1];while (i + j <= n && tmp + j <= n && s[i+j] == s[tmp+j]) ++j;height[Rank[i]] = j;} } int check(int x) {int minsa = 1e9, maxsa = -1e9;for (int i = 1; i <= n; ++i) {if (height[i] >= x) {minsa = min(minsa, Sa[i]);maxsa = max(maxsa, Sa[i]);if (maxsa - minsa >= x) return 1;}else {minsa = Sa[i];maxsa = Sa[i];}}return 0; }int main () {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);cin >> n;for (int i = 1; i <= n; ++i) {cin >> s[i];}suffixArray();int l = 1, r = n;while (l <= r) {if (check(mid)) l = mid + 1;else r = mid - 1;} cout << r << endl;return 0; }

總結

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