POJ - 3261 Milk Patterns (尾碼數組求可重疊的 k 次最長重複子串)

標籤：des   style   blog   http   color   io   os   ar   for   

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

題意:給定一個字串，求至少出現k 次的最長重複子串，這k個子串可以重疊。
思路：和上一題POJ - 1743 Musical Theme一樣也是二分答案，然後將尾碼分成若干組，不同的是，這裡要判斷的是有沒有一個組的尾碼heigh[]不小於mid。如果有，那麼再有連續的k個存在的話，就滿足條件。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 20010;int sa[maxn]; //SA數組，表示將S的n個尾碼從小到大排序後把排好序的//的尾碼的開頭位置順次放入SA中int t1[maxn], t2[maxn], c[maxn];int rank[maxn], height[maxn];int s[maxn];void build_sa(int s[], int n, int m) {    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] = 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 (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 (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]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+j] == y[sa[i]+j] ? p-1 : p++;        if (p >= n) break;        m = p;    }}void getHeight(int s[],int n) {    int i, j, k = 0;    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 (s[i+k] == s[j+k]) k++;        height[rank[i]]=k;    }}int check(int n, int k, int mid) {    int num = 1;    for (int i = 2; i <= n; i++) {        if (height[i] >= mid) {            num++;            if (num >= k) return 1;        }        else num = 1;    }    return 0;}int main() {    int n, k;    while (scanf("%d%d", &n, &k) != EOF) {        int Max = 0;        for (int i = 0; i < n; i++) {            scanf("%d", &s[i]);            Max = max(Max, s[i]);        }        s[n] = 0;  //notice        build_sa(s, n+1, Max+1);        getHeight(s, n);        int l = 0, r = n;        int ans = 0;        while (l <= r) {            int mid = l + r >> 1;            if (check(n, k, mid)) {                ans = mid;                l = mid + 1;            }            else r = mid - 1;        }        printf("%d\n", ans);    }    return 0;}

POJ - 3261 Milk Patterns (尾碼數組求可重疊的 k 次最長重複子串)