POJ 3261 Milk Patterns （后缀树组）

Description

Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.

Input

Line 1: Two space-separated integers: N and K 
Lines 2..N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at least K times

Sample Input

8 212323231

Sample Output

4

题目大意：给一个长度为n的串，求最少出现k次的最长子串（可以相互重叠）。

思路：与POJ1743相似。先二分答案length，将排好序后的后缀按length分成若干组，每组内的height值都不小于length。若该组的后缀个数不小于k，则满足条件。

#include<cstdio>#include<cstring>#include<vector>#include<set>#include<queue>#include<map>#include<iostream>#include<stack>#include<cctype>#include<algorithm>using namespace std;const int maxn = 20000 + 10;int a[maxn], b[maxn];int t1[maxn], t2[maxn], c[maxn];int Rank[maxn], height[maxn];int r[maxn], sa[maxn];bool cmp(int *r, int a, int b, int l){    return r[a] == r[b] && r[a+l] == r[b+l];}void da(int str[], int sa[], int Rank[], int height[], int n, int m){    n++;    int i, j, p, *x=t1, *y=t2;    for(i = 0; i < m; i++) c[i] = 0;    for(i = 0; i < n; i++) c[ x[i] = str[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(j = 1; j <= n; j <<= 1)    {        p = 0;        for(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(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(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], j) ? p-1 : p++;        if(p >= n) break;        m = p;    }    int k = 0;    n--;    for(i = 0; i <= n; i++) Rank[sa[i]] = i;    for(i = 0; i < n; i++)    {        if(k) k--;        j = sa[Rank[i]-1];        while(str[i+k] == str[j+k]) k++;        height[Rank[i]] = k;    }}int RMQ[maxn], mm[maxn], best[20][maxn];void initRMQ(int n){    mm[0] = -1;    for(int i = 1; i <= n; i++)        mm[i] = ((i&(i-1))==0) ? mm[i-1]+1 : mm[i-1];    for(int i = 1; i <= n; i++) best[0][i] = i;    for(int i = 1; i <= mm[n]; i++)        for(int j = 1; j+(1<<i)-1 <= n; j++)        {            int a = best[i-1][j];            int b = best[i-1][j + (1<<(i-1))];            if(RMQ[a] < RMQ[b]) best[i][j] = a;            else best[i][j] = b;        }}int askRMQ(int a, int b){    int t;    t = mm[b-a+1];    b -= (1<<t)-1;    a = best[t][a]; b = best[t][b];    return RMQ[a] < RMQ[b] ? a : b;}int lcp(int a, int b){    a = Rank[a], b = Rank[b];    if(a > b) swap(a, b);    return height[askRMQ(a+1, b)];}bool judge(int k, int n){int mi, mx;mi = mx = sa[1];for(int i = 2; i <= n; i++) {if(height[i] < k) mx = mi = sa[i];else {mx = max(mx, sa[i]);mi = min(mi, sa[i]);if(mx - mi >= k) return true;}}return false;}bool judge(int l, int k, int n){    int cnt = 0;    for(int i = 2;  i <= n; i++)     {    if(height[i] < l) cnt = 0;    else {    cnt++;    if(cnt+1 >= k) return true;    }    }    return false;}int main(){    int n, k;    scanf("%d%d", &n, &k);    for(int i = 0; i < n; i++) { scanf("%d", &a[i]); b[i] = a[i]; }    sort(b, b+n); int len = unique(b, b+n) - b;    for(int i = 0; i < n; i++) {    a[i] = upper_bound(b, b+len, a[i]) - b;    }    a[n] = 0;    da(a, sa, Rank, height, n, 20005);    int L = 0, R = n, ans = 0;    while(L <= R)    {    int mid = (L + R) >> 1;    if(judge(mid, k, n)) { ans = mid; L = mid + 1; }    else R = mid - 1;    }    cout << ans << endl;    return 0;}