Codeforces Round #246 (Div. 2) D. Prefixes and Suffixe 后缀数组

链接：

codeforces.com/contest/432/problem/D

题意：

给你一个字符串，求每一个和前缀匹配的后缀在这个字符串中出现的次数

题解：

先算出lcp，找到sa[i]==0的位置标记为beg，和前缀匹配的后缀一定会出现beg的左边，这个想一想明白了

我们先算出beg左边每一个后缀和beg匹配的长度，beg右边的先放到一个vector中，便于之后二分查找

我们从beg向左遍历，对于每一个dp[i]，如果dp[i]==n-sa[i]，就可以更新结果了，从i到beg中间的都满足，同时还要加上beg右边的

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 dp[MAXN];

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    string s;
    cin >> s;
    construct_sa(s, sa);
    construct_lcp(s, sa, lcp);
    int beg;
    rep(i, 0, n + 1) if (sa[i] == 0) {
        beg = i;
        break;
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[beg] = INF;
    per(i, 0, beg) dp[i] = min(dp[i + 1], lcp[i]);
    rep(i, beg + 1, n + 1) dp[i] = min(dp[i - 1], lcp[i - 1]);
    VI v;
    rep(i, beg + 1, n + 1) if (dp[i]) v.pb(dp[i]);
    sort(all(v));
    map<int, int> mmp;
    per(i, 0, beg) {
        if (!dp[i]) break;
        if (dp[i] == n - sa[i]) {
            mmp[dp[i]] += beg - i + 1;
            mmp[dp[i]] += v.end() - lower_bound(all(v), dp[i]);
        }
    }
    cout << mmp.size() + 1 << endl;
    for (auto it : mmp) cout << it.first << ' ' << it.second << endl;
    cout << n << ' ' << 1 << endl;
    return 0;
}