思路

將兩個字符串合成一個字符串，中間加上一個"#"。然后求$height[i]$。
最長公共子串如果存在，必定是相鄰的$height[i]$，求最大值的時候判斷是否是來自兩個不同的子串。

#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 = 2e5 + 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]; char s[maxn]; int n; void SuffixArray () {for (int i = 0; i < 1000; ++i) cntA[i] = 0;for (int i = 1; i <= n; ++i) cntA[(int)s[i]]++;for (int i = 1; i < 1000; ++i) cntA[i] += cntA[i-1];for (int i = n; i >= 1; --i) Sa[ cntA[(int)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 pos; int check (int a, int b) {if (a < pos && b > pos) return 0;return 1; }int main () {ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);s[0] = ' ';scanf("%s", s+1);int len = strlen(s);s[len] = '#';pos = len;scanf("%s", s + len + 1);n = strlen(s) - 1;SuffixArray();int ans = 0;for (int i = 1; i <= n; ++i) {int a = Sa[i];int b = Sa[i - 1];if (a > b) swap(a, b);if (height[i] <= ans || check(a, b)) continue;ans = height[i];// ans = max(ans, min(height[i], pos - a));}cout << ans << endl;return 0; }

總結

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