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：

int n, k;
int Rank[MAXN], tmp[MAXN];
int sa[MAXN], lcp[MAXN];

bool compare_sa(int i, int j) {
    if (Rank[i] != Rank[j]) return Rank[i] < Rank[j];
    else {
        int ri = i + k <= n ? Rank[i + k] : -1;
        int rj = j + k <= n ? Rank[j + k] : -1;
        return ri < rj;
    }
}

void construct_sa(string S, int *sa) {
    n = S.length();
    for (int i = 0; i <= n; i++) {
        sa[i] = i;
        Rank[i] = i < n ? S[i] : -1;
    }
    for (k = 1; k <= n; k *= 2) {
        sort(sa, sa + n + 1, compare_sa);
        tmp[sa[0]] = 0;
        for (int i = 1; i <= n; i++)
            tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
        for (int i = 0; i <= n; i++) Rank[i] = tmp[i];
    }
}

void construct_lcp(string S, int *sa, int *lcp) {
    int n = S.length();
    for (int i = 0; i <= n; i++) Rank[sa[i]] = i;
    int h = 0;
    lcp[0] = 0;
    for (int i = 0; i < n; i++) {
        int j = sa[Rank[i] - 1];
        if (h > 0) h--;
        for (; j + h < n && i + h < n; h++)
            if (S[j + h] != S[i + h]) break;
        lcp[Rank[i] - 1] = h;
    }
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    string s1, s2;
    cin >> s1 >> s2;
    int n1 = s1.length();
    string s = s1 + "$" + s2;
    construct_sa(s, sa);
    construct_lcp(s, sa, lcp);
    int ans = INF;
    rep(i, 1, n + 1) if (lcp[i] > lcp[i - 1] && lcp[i] > lcp[i + 1])
        if (!((sa[i] > n1) ^ (sa[i + 1] < n1)))
            ans = min(ans, max(lcp[i - 1], lcp[i + 1]) + 1);
    if (ans == INF) ans = -1;
    cout << ans << endl;
    return 0;
}