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