POJ 3261 (后缀数组 二分) Milk Patterns

这道题和UVa 12206一样，求至少重复出现k次的最长字串。

首先还是二分最长字串的长度len，然后以len为边界对height数组分段，如果有一段包含超过k个后缀则符合要求。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <algorithm>

 4 using namespace std;

 5 

 6 const int maxn = 20000 + 10;

 7 const int maxm = 1000000 + 10;

 8 

 9 int s[maxn];

10 int sa[maxn], height[maxn], rank[maxn];

11 int t[maxn], t2[maxn], c[maxm];

12 int n, k;

13 

14 void build_sa(int m)

15 {

16     int i, *x = t, *y = t2;

17     for(i = 0; i < m; i++) c[i] = 0;

18     for(i = 0; i < n; i++) c[x[i] = s[i]]++;

19     for(i = 1; i < m; i++) c[i] += c[i - 1];

20     for(i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;

21     for(int k = 1; k <= n; k <<= 1)

22     {

23         int p = 0;

24         for(i = n - k; i < n; i++) y[p++] = i;

25         for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;

26         for(i = 0; i < m; i++) c[i] = 0;

27         for(i = 0; i < n; i++) c[x[y[i]]]++;

28         for(i = 1; i < m; i++) c[i] += c[i - 1];

29         for(i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];

30         swap(x, y);

31         p = 1; x[sa[0]] = 0;

32         for(i = 1; i < n; i++)

33             x[sa[i]] = y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k] ? p - 1 : p++;

34         if(p >= n) break;

35         m = p;

36     }

37 }

38 

39 void build_height()

40 {

41     int i, j, k = 0;

42     for(i = 0; i < n; i++) rank[sa[i]] = i;

43     for(i = 0; i < n; i++)

44     {

45         if(k) k--;

46         j = sa[rank[i] - 1];

47         while(s[i + k] == s[j + k]) k++;

48         height[rank[i]] = k;

49     }

50 }

51 

52 bool ok(int len)

53 {

54     int cnt = 0;

55     for(int i = 0; i < n; i++)

56     {

57         if(i == 0 || height[i] < len) cnt = 0;

58         if(++cnt >= k) return true;

59     }

60     return false;

61 }

62 

63 int main()

64 {

65     //freopen("in.txt", "r", stdin);

66 

67     scanf("%d%d", &n, &k);

68     for(int i = 0; i < n; i++)

69     {

70         scanf("%d", &s[i]);

71         s[i]++;

72     }

73     s[n] = 0;

74     build_sa(1000002);

75     build_height();

76 

77     int L = 1, R = n;

78     while(L < R)

79     {

80         int M = (L + R + 1) / 2;

81         if(ok(M)) L = M;

82         else R = M - 1;

83     }

84 

85     printf("%d\n", L);

86 

87     return 0;

88 }

代码君