Codeforces Round #244 (Div. 2) D. Match & Catch 后缀数组

链接：

http://codeforces.com/contest/427/problem/D

题意：

给你两个字符串s1,s2，找出最短的子串出现在s1和s2中有且只有一次

题解：

还是把s1和s2连起来，求lcp。首先要知道得是，最短长度一定是sa数组中一定是相连的，

这样就只需要遍历一遍lcp数组，更新ans就可以了

ans = min(ans, max(lcp[i - 1], lcp[i + 1]) + 1)

代码：

31 int n, k;
32 int Rank[MAXN], tmp[MAXN];
33 int sa[MAXN], lcp[MAXN];
34 
35 bool compare_sa(int i, int j) {
36     if (Rank[i] != Rank[j]) return Rank[i] < Rank[j];
37     else {
38         int ri = i + k <= n ? Rank[i + k] : -1;
39         int rj = j + k <= n ? Rank[j + k] : -1;
40         return ri < rj;
41     }
42 }
43 
44 void construct_sa(string S, int *sa) {
45     n = S.length();
46     for (int i = 0; i <= n; i++) {
47         sa[i] = i;
48         Rank[i] = i < n ? S[i] : -1;
49     }
50     for (k = 1; k <= n; k *= 2) {
51         sort(sa, sa + n + 1, compare_sa);
52         tmp[sa[0]] = 0;
53         for (int i = 1; i <= n; i++)
54             tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
55         for (int i = 0; i <= n; i++) Rank[i] = tmp[i];
56     }
57 }
58 
59 void construct_lcp(string S, int *sa, int *lcp) {
60     int n = S.length();
61     for (int i = 0; i <= n; i++) Rank[sa[i]] = i;
62     int h = 0;
63     lcp[0] = 0;
64     for (int i = 0; i < n; i++) {
65         int j = sa[Rank[i] - 1];
66         if (h > 0) h--;
67         for (; j + h < n && i + h < n; h++)
68             if (S[j + h] != S[i + h]) break;
69         lcp[Rank[i] - 1] = h;
70     }
71 }
72 
73 int main() {
74     ios::sync_with_stdio(false), cin.tie(0);
75     string s1, s2;
76     cin >> s1 >> s2;
77     int n1 = s1.length();
78     string s = s1 + "$" + s2;
79     construct_sa(s, sa);
80     construct_lcp(s, sa, lcp);
81     int ans = INF;
82     rep(i, 1, n + 1) if (lcp[i] > lcp[i - 1] && lcp[i] > lcp[i + 1])
83         if (!((sa[i] > n1) ^ (sa[i + 1] < n1)))
84             ans = min(ans, max(lcp[i - 1], lcp[i + 1]) + 1);
85     if (ans == INF) ans = -1;
86     cout << ans << endl;
87     return 0;
88 }