BZOJ1717: [Usaco2006 Dec]Milk Patterns 产奶的模式

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 20000+1000
14 #define maxm 1000000+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 using namespace std;
19 inline int read()
20 {
21     int x=0,f=1;char ch=getchar();
22     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
23     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
24     return x*f;
25 }
26 int wa[maxn],wb[maxn],wr[maxm],a[maxn],sa[maxn],rank[maxn],h[maxn],n,k;
27 inline bool cmp(int *r,int a,int b,int l)
28 {
29     return r[a]==r[b]&&r[a+l]==r[b+l];
30 }
31 void da(int *r,int *sa,int n,int m)
32 {
33     int i,j,p,*x=wa,*y=wb,*t;
34     for(i=0;i<m;i++)wr[i]=0;
35     for(i=0;i<n;i++)wr[x[i]=r[i]]++;
36     for(i=1;i<m;i++)wr[i]+=wr[i-1];
37     for(i=n-1;i>=0;i--)sa[--wr[x[i]]]=i;
38     for(j=1,p=1;p<n;j<<=1,m=p)
39      {
40          for(p=0,i=n-j;i<n;i++)y[p++]=i;
41          for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;
42          for(i=0;i<m;i++)wr[i]=0;
43          for(i=0;i<n;i++)wr[x[y[i]]]++;
44          for(i=1;i<m;i++)wr[i]+=wr[i-1];
45          for(i=n-1;i>=0;i--)sa[--wr[x[y[i]]]]=y[i];
46          for(t=x,x=y,y=t,p=1,i=1,x[sa[0]]=0;i<n;i++)
47           x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
48      }
49 }
50 void geth(int *r,int *sa,int n)
51 {
52     int i,j,k=0;
53     for(i=1;i<=n;i++)rank[sa[i]]=i;
54     for(i=0;i<n;h[rank[i++]]=k)
55     for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);
56 }
57 bool check(int x)
58 {
59     int cnt=1;
60     for(int i=1;i<=n;i++)
61      if(h[i]>=x)
62       {
63           cnt++;
64           if(cnt>=k)return 1;
65       }
66      else cnt=1; 
67     return 0; 
68 }
69 int main()
70 {
71     freopen("input.txt","r",stdin);
72     freopen("output.txt","w",stdout);
73     n=read();k=read();
74     for(int i=0;i<n;i++)a[i]=read()+1;
75     a[n]=0;
76     da(a,sa,n+1,1000002);
77     geth(a,sa,n);
78     int l=1,r=n,mid;
79     while(l<=r)
80     {
81         mid=(l+r)>>1;
82         if(check(mid))l=mid+1;else r=mid-1;
83     }
84     printf("%d\n",r);
85     return 0;
86 }