poj 3261 Milk Patterns (最长公共子串，后缀数组)
题意：给一个字符串，要求找出至少出现k次的最长重复子串...
思路：后缀数组，然后再次用到了根据height数组对后缀进行分组的套路...二分判定合法性，对于当前的最长长度x,分组使得每组中的height[i]都大于等于x,所不同的是，判定变成了存在一个组，后缀的个数至少为k个（因为这样，就可以对于大于等于k个的后缀，同时取前x长度，得到的就是出现了至少k次且长度为x的前缀）1A,蛤蛤蛤
1/* ***********************************************
2Author :111qqz
3Created Time :2016年08月01日 星期一 01时30分34秒
4File Name :code/poj/3261.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31const int N=2E4+7;
32const int M=2E6+11;
33const int C = 5;
34int n,sa[N],rk[N],t[N],t2[N],cnt[M];
35int height[N];
36int s[N];
37int k;
38int cmp(int *r,int a,int b,int l){ return r[a]==r[b]&&r[a+l]==r[b+l];}
39void getSa( int n,int m)
40{
41 int *x = t;
42 int *y = t2;
43 ms(cnt,0);
44 for ( int i = 0 ; i < n; i++) cnt[x[i]=s[i]]++;
45 for ( int i = 1 ; i < m ; i++) cnt[i]+=cnt[i-1];
46 for ( int i = n-1 ; i >= 0 ; i--) sa[--cnt[x[i]]] = i;
47 for ( int k = 1 ; k <= n ; k<<=1)
48 {
49 int p = 0;
50 for ( int i = n-k ; i < n; i++) y[p++] = i ;
51 for ( int i = 0 ; i < n; i++) if (sa[i]>=k) y[p++] = sa[i]-k;
52 ms(cnt,0);
53 for ( int i = 0 ; i <n ; i++) cnt[x[y[i]]]++;
54 for ( int i = 0 ; i <m ; i++) cnt[i]+=cnt[i-1];
55 for ( int i = n-1 ; i >= 0 ; i--) sa[--cnt[x[y[i]]]] = y[i];
56 swap(x,y);
57 p = 1;
58 x[sa[0]] = 0 ;
59 for ( int i = 1 ; i < n; i++)
60 x[sa[i]] = cmp(y,sa[i-1],sa[i],k)?p-1:p++;
61 if (p>=n) break;
62 m = p;
63 }
64}
65void getHeight( int n)
66{
67 int k = 0 ;
68 for ( int i = 0 ; i < n; i++) rk[sa[i]] = i ;
69 height[0] = 0;
70 for ( int i = 0 ; i < n; i ++)
71 {
72 if (rk[i]==0) continue;
73 if (k) k--;
74 int j = sa[rk[i]-1];
75 while (s[i+k]==s[j+k]) k++;
76 height[rk[i]] = k ;
77 }
78}
79int getSuffix( int s[],int len)
80{
81 int up = 0 ;
82 for ( int i = 0 ; i < len ; i++)
83 {
84 up = max(up,s[i]);
85 }
86 s[len++] = 0 ;
87 getSa(len,up+1);
88 getHeight(len);
89 return len;
90}
91bool check( int x,int k,int n)
92{
93 int cnt = 1 ;
94 for ( int i = 2; i <= n ; i++)
95 {
96 if (height[i]>=x&&i<n)
97 {
98 cnt++;
99 continue;
100 }
101 if (cnt>=k) return true;
102 cnt = 1;
103 }
104 return false;
105}
106int main()
107{
108 #ifndef ONLINE_JUDGE
109 freopen("code/in.txt","r",stdin);
110 #endif
111 scanf("%d %d",&n,&k);
112 for ( int i = 0 ; i < n ; i++) scanf("%d",&s[i]),s[i]+=C;
113 n = getSuffix(s,n);
114// for ( int i = 0 ; i < n; i++) cout<<"s[i]:"<<s[i]<<endl;
115 int l = 1,r = n;
116 int ans = 0 ;
117 while (l<=r)
118 {
119 int mid = (l+r)/2;
120 if (check(mid,k,n))
121 {
122 ans = mid;
123 l = mid + 1;
124 }
125 else r = mid-1;
126 }
127 printf("%d\n",ans);
128 #ifndef ONLINE_JUDGE
129 fclose(stdin);
130 #endif
131 return 0;
132}