后缀数组 POJ 3261 Milk Patterns

题目链接

题意：可重叠的 k 次最长重复子串。给定一个字符串，求至少出现 k 次的最长重复子串，这 k 个子串可以重叠。

分析：与POJ 1743做法类似，先二分答案，height数组分段后统计 LCP>=m 的子串的个数。

#include <cstdio>
#include <cstring>
#include <algorithm>

const int N = 2e4 + 5;
int sa[N], rank[N], height[N];
int t[N], t2[N], c[N];
int a[N], A[N];

void da(int *s, int n, int m) {
    int i, p, *x = t, *y = t2;
    for (i=0; i<m; ++i) c[i] = 0;
    for (i=0; i<n; ++i) c[x[i]=s[i]]++;
    for (i=1; i<m; ++i) c[i] += c[i-1];
    for (i=n-1; i>=0; --i) sa[--c[x[i]]] = i;
    for (int k=1; k<=n; k<<=1) {
        for (p=0, i=n-k; i<n; ++i) y[p++] = i;
        for (i=0; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;
        for (i=0; i<m; ++i) c[i] = 0;
        for (i=0; i<n; ++i) c[x[y[i]]]++;
        for (i=0; i<m; ++i) c[i] += c[i-1];
        for (i=n-1; i>=0; --i) sa[--c[x[y[i]]]] = y[i];
        std::swap (x, y);
        p = 1; x[sa[0]] = 0;
        for (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;
    }
}

void calc_height(int n) {
    int i, k = 0;
    for (i=0; i<n; ++i) rank[sa[i]] = i;
    for (i=0; i<n; ++i) {
        if (k) k--;
        int j = sa[rank[i]-1];
        while (a[i+k] == a[j+k]) k++;
        height[rank[i]] = k;
    }
}

int n, k;

bool check(int m) {
    int cnt = 0;
    for (int i=1; i<n; ++i) {
        if (height[i] >= m) {
            cnt++;
            //height[i] = LCP (suffix (sa[i-1], sa[i]));
            if (cnt + 1 >= k) {
                return true;
            }
        } else {
            cnt = 0;
        }
    }
    return false;
}

int main() {
    while (scanf ("%d%d", &n, &k) == 2) {
        for (int i=0; i<n; ++i) {
            scanf ("%d", a+i);
            A[i] = a[i];
        }
        std::sort (A, A+n);
        for (int i=0; i<n; ++i) {
            a[i] = std::lower_bound (A, A+n, a[i]) - A + 1;
        }
        a[n++] = 0;

        da (a, n, 20000);
        calc_height (n);

        int ans = 0;
        int left = 0, right = n - 1;
        while (left <= right) {
            int mid = left + right >> 1;
            if (check (mid)) {
                ans = std::max (ans, mid);
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        printf ("%d\n", ans);
    }
    return 0;
}