bzoj 1717 [Usaco2006 Dec]Milk Patterns 产奶的模式 后缀数组

题面

题目传送门

解法

比较套路的一道题吧

先求出height数组，二分答案\(mid\)

将后缀分成若干组，每一组满足height≥mid

若某一组的后缀个数≥k，那么答案可行

时间复杂度：\(O(n\ log\ n)\)

代码

#include <bits/stdc++.h>
#define N 1000010
using namespace std;
template <typename node> void read(node &x) {
    x = 0; int f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
    while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
int n, k, st[N];
struct SuffixArray {
    int x[N], y[N], sa[N], rnk[N], cnt[N], height[N];
    void build() {
        int m = 0;
        for (int i = 1; i <= n; i++) m = max(m, st[i]);
        for (int i = 1; i <= n; i++) cnt[x[i] = st[i]]++;
        for (int i = 2; i <= m; i++) cnt[i] += cnt[i - 1];
        for (int i = n; i; i--) sa[cnt[x[i]]--] = i;
        for (int k = 1; k <= n; k <<= 1) {
            int num = 0;
            for (int i = n - k + 1; i <= n; i++) y[++num] = i;
            for (int i = 1; i <= n; i++)
                if (sa[i] > k) y[++num] = sa[i] - k;
            for (int i = 1; i <= m; i++) cnt[i] = 0;
            for (int i = 1; i <= n; i++) cnt[x[i]]++;
            for (int i = 1; i <= m; i++) cnt[i] += cnt[i - 1];
            for (int i = n; i; i--) sa[cnt[x[y[i]]]--] = y[i], y[i] = 0;
            swap(x, y); x[sa[1]] = 1, num = 1;
            for (int i = 2; i <= n; i++)
                x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? num : ++num;
            if (num == n) break; m = num;
        }
        for (int i = 1; i <= n; i++) rnk[sa[i]] = i;
        int k = 0;
        for (int i = 1; i <= n; i++) {
            if (rnk[i] == 1) continue;
            if (k) k--; int j = sa[rnk[i] - 1];
            while (i + k <= n && j + k <= n && st[i + k] == st[j + k]) k++;
            height[rnk[i]] = k;
        }
    }
    bool check(int mid) {
        int las = 1;
        for (int i = 2; i <= n; i++)
            if (height[i] < mid) {
                if (i - las >= k) return true;
                las = i;
            }
        return (n - las + 1) >= k;
    }
    int solve() {
        int l = 0, r = n, ans = 0;
        while (l <= r) {
            int mid = (l + r) >> 1;
            if (check(mid)) ans = mid, l = mid + 1;
                else r = mid - 1;
        }
        return ans;
    }
} SA;
int main() {
    read(n), read(k);
    for (int i = 1; i <= n; i++) read(st[i]);
    SA.build();
    cout << SA.solve() << "\n";
    return 0;
}