SPOJ 694/705 Distinct Substrings 後綴數組經典應用

問題本質

求字符串本質不同子串數量。子串必是某後綴的前綴。後綴數組 SA 排序後，第 i 個後綴貢獻 n - SA[i] - height[i] 個新子串（扣除與前一後綴的公共前綴）。

後綴數組構建（倍增法）

void suffix(int m) {
    int *x = t1, *y = t2;
    for (int i = 0; i < m; i++) c[i] = 0;
    for (int i = 0; i < n; i++) c[x[i] = str[i]]++;
    for (int i = 1; i < m; i++) c[i] += c[i-1];
    for (int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
    for (int k = 1; k <= n; k <<= 1) {
        int p = 0;
        for (int i = n-k; i < n; i++) y[p++] = i;
        for (int i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;
        for (int i = 0; i < m; i++) c[i] = 0;
        for (int i = 0; i < n; i++) c[x[y[i]]]++;
        for (int i = 1; i < m; i++) c[i] += c[i-1];
        for (int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for (int i = 1; i < n; i++)
            x[sa[i]] = (y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k]) ? p-1 : p++;
        if (p >= n) break;
        m = p;
    }
}

Height 數組計算

void getheight() {
    int k = 0;
    for (int i = 0; i < n; i++) rank[sa[i]] = i;
    for (int i = 0; i < n; i++) {
        if (k) k--;
        if (!rank[i]) continue;
        int j = sa[rank[i]-1];
        while (str[i+k] == str[j+k]) k++;
        height[rank[i]] = k;
    }
}

答案計算

int ans = 0;
for (int i = 1; i < n; i++) ans += n - 1 - sa[i] - height[i];

時間複雜度 O(n log n)，適用於 10⁵ 級字符串。注意字符串末尾添加哨兵 0。