POJ 3415 后缀数组+单调栈

链接：

http://poj.org/problem?id=3415

题意：

统计A和B长度不小于K的公共子串个数。

题解：

将A和B拼接后，利用单调栈累计分属两者的后缀对应的LCP-K+1即为答案

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;
    }
}

PII st[MAXN];
ll contribution, top;

ll solve(int k, int n1, bool is_s1) {
    ll ans = 0;
    rep(i, 0, n) {
        if (lcp[i] < k) {
            top = contribution = 0;
            continue;
        }
        int size = 0;
        if (is_s1 && sa[i] < n1 || !is_s1 && sa[i] > n1) {
            size++;
            contribution += lcp[i] - k + 1;
        }
        while (top > 0 && st[top].first >= lcp[i]) {
            contribution -= st[top].second*(st[top].first - lcp[i]);
            size += st[top].second;
            top--;
        }
        st[++top] = mp(lcp[i], size);
        if (is_s1 && sa[i + 1] > n1 || !is_s1 && sa[i + 1] < n1) ans += contribution;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    int k;
    while (cin >> k, k) {
        string a, b;
        cin >> a >> b;
        int n1 = a.length();
        string s = a + '$' + b;
        construct_sa(s, sa);
        construct_lcp(s, sa, lcp);
        cout << solve(k, n1, true) + solve(k, n1, false) << endl;
    }
    return 0;
}