最长不重叠子串

很容易就会让人想到后缀数组。我们前后做一个差，然后就是类似求最长不重叠子串了。
若两后缀的相同前缀长为i，那么最后就有i+1长的主题。
所以这里不光是不重叠，还要空一。
如何求解最长不重叠子串？二分，对height数组进行分组。(非常的巧妙！建议百度)

代码如下：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<math.h>
#include<cstdio>
using namespace std;
const int max_n = 2e4 + 100;
int a[max_n];

//后缀数组模板。自己写的太慢了呜呜
int ranks[max_n], SA[max_n], height[max_n];
int wa[max_n], wb[max_n], wvarr[max_n], wsarr[max_n];
inline int cmp(int* r, int a, int b, int l) {
    return r[a] == r[b] && r[a + l] == r[b + l];
}
inline void get_sa(int* r, int* sa, int n, int m) {
    //r:原数组,我们此处默认为a
    //sa:SA
    //n:原数组长度
    //m:原数组种类数，用于基数排序
    ++n;
    int i, j, p, * x = wa, * y = wb, * t;
    for (i = 0; i < m; ++i) wsarr[i] = 0;
    for (i = 0; i < n; ++i) wsarr[x[i] = r[i]]++;
    for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1];
    for (i = n - 1; i >= 0; --i) sa[--wsarr[x[i]]] = i;
    for (j = 1, p = 1; p < n; j <<= 1, m = p) {
        for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
        for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
        for (i = 0; i < n; ++i) wvarr[i] = x[y[i]];
        for (i = 0; i < m; ++i) wsarr[i] = 0;
        for (i = 0; i < n; ++i) wsarr[wvarr[i]]++;
        for (i = 1; i < m; ++i) wsarr[i] += wsarr[i - 1];
        for (i = n - 1; i >= 0; --i) sa[--wsarr[wvarr[i]]] = y[i];
        for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}
//求解高度数组，height[i]指排名
void get_height(int* r, int* sa, int n) {
    int i, j, k = 0;
    for (i = 1; i <= n; ++i) ranks[sa[i]] = i;
    for (i = 0; i < n; height[ranks[i++]] = k)
        for (k ? k-- : 0, j = sa[ranks[i] - 1]; r[i + k] == r[j + k]; k++);
    return;
}
void init(int n) {
    fill(a, a + n + 3, 0);
    fill(ranks, ranks + n + 3, 0);
    fill(SA, SA + n + 3, 0);
    fill(height, height + n + 3, 0);
    fill(wa, wa + n + 3, 0);
    fill(wb, wb + n + 3, 0);
    fill(wsarr, wsarr + n + 3, 0);
    fill(wvarr, wvarr + n + 3, 0);
}
//后缀数组

//附加rmq_st 专门对后缀数组
int st[max_n][32];
void initSt(int n) {
    for (int i = 0;i <= n;++i)st[i][0] = height[i];
    int mxk = (int)log2(n + 1);
    for (int k = 1;k <= mxk;++k) {
        for (int i = 0;i <= n;++i) {
            if (i + (1 << k) - 1 > n)break;
            st[i][k] = min(st[i][k - 1], st[i + (1 << (k - 1))][k - 1]);
        }
    }
}
int que(int l, int r) {
    l = ranks[l];r = ranks[r];
    if (l > r)swap(l, r);
    ++l;
    int len = log2(r - l + 1);
    return min(st[l][len], st[r - (1 << len) + 1][len]);
}
int n;
bool check(int len) {
    for (int i = 1, mi, ma;i <= n;++i) {
        if (height[i] < len - 1) {
            mi = SA[i];ma = SA[i];
        }
        else {
            mi = min(min(SA[i], mi), SA[i - 1]);
            ma = max(max(SA[i], ma), SA[i - 1]);
            if (ma - mi >= len)return true;
        }
    }return false;
}
int ef() {
    int lft = 0;int rght = n + 1;
    while (lft < rght) {
        int mid = (lft + rght) >> 1;
        if (check(mid))lft = mid + 1;
        else rght = mid;
    }return rght - 1;
}
int main() {
    while (scanf("%d", &n)) {
        if (n == 0)break;
        init(n);
        for (int i = 0;i < n;++i)scanf("%d", &a[i]);
        for (int i = 0;i < n - 1;++i)a[i] = a[i + 1] - a[i];
        for (int i = 0;i < n;++i)a[i] += 90;
        get_sa(a, SA, n, 300);
        get_height(a, SA, n);
        initSt(n);int ans = ef();
        printf("%d\n", ans > 4 ? ans : 0);
    }
}