hihocoder 第120周 后缀数组 + 单调队列

分析：

求出高度数组，用单调队列维护区间最小值。高度数组的区间最小值的意义就是这些排名相邻的后缀的最长公共前缀。实际上是把所有字符串划分为后缀。这里的是可重叠子串。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define pr(x) cout << #x << ": " << x << "  " 
#define pl(x) cout << #x << ": " << x << endl;

static const int maxn = int(2e4) + 13;
//后缀数组模板，时间复杂度nlogn,空间复杂度n
struct Suffix
{

    int r[maxn];
    //后缀数组，名词数组，高度数组（表示和它前一名后缀的最长前缀长度, 从2开始有意义)。
    int sa[maxn],rnk[maxn],height[maxn];
    int t[maxn],t2[maxn],c[maxn],n,m;

    char str[maxn];

    //把string转化为r[i]
    void init_string(string &s) {
        n = s.size();
        for(int i=0;i<n;i++) r[i]=(int)s[i];
        m = 128;
    }

    void init_char(char *s) {
        n = strlen(s);
        for(int i=0;i<n;i++) r[i]=(int)s[i];
        m = 128;
    }

    int cmp(int *r,int a,int b,int l) { return r[a]==r[b]&&r[a+l]==r[b+l]; }

    //处理得到后缀数组
    void build() {
        int i,k,p,*x=t,*y=t2;
        r[n++]=0;
        for (i=0; i<m; i++) c[i]=0;
        for (i=0; i<n; i++) c[x[i]=r[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 (k=1,p=1; k<n; k*=2,m=p) {
            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=1; i<m; i++) c[i]+=c[i-1];
            for (i=n-1; i>=0; i--) sa[--c[x[y[i]]]]=y[i];
            swap(x,y);
            p=1;
            x[sa[0]]=0;
            for (i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],k)?p-1:p++;
        }
        n--;
    }

    //处理得到高度数组和名次数组
    void lcp() {
        int i, j, k = 0;
        for (i=1; i <= n; i++) rnk[sa[i]]=i;
        for (i=0; i < n; i++) {
            if (k) k--;
            j = sa[rnk[i] - 1];
            while (r[i + k]==r[j + k]) k++;
            height[rnk[i]] = k;
        }
    }
};

struct jibancanyang
{

    int que[maxn];

    void fun() {
        int n, k;
        scanf("%d%d", &n, &k);
        string str;
        for (int i = 0; i < n; ++i) {
            int x;
            scanf("%d", &x);
            str += char(x + '0');
        }
        if (k == 1) {
            printf("%d\n", n);
            return;
        }
        Suffix now;
        now.init_string(str);
        now.build();
        now.lcp();
        int head = 0, tail = 0;
        int l = 2, r = 2, ans = 0;
        while (r <= n) {
            while (r - l < k - 1) {
                if (r > n) break;
                int x = now.height[r++];
                while (tail != head && x < que[tail - 1]) --tail;
                que[tail++] = x;
            }
            if (r > n) break;
            ans = max(ans, que[head]);
            if (que[head] == now.height[l++]) head++;
        }
        printf("%d\n", ans);
    }

}ac;

int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif
    ac.fun();
    return 0;
}