bzoj1717(poj3261) [Usaco2006 Dec]Milk Patterns 产奶的模式(后缀数组，二分答案)

#include <cstdio>
#include <cstring>
#include <algorithm>
#define  N 20005
int n,m,a
,aa
;
int rank[N<<1],count
,rank1
,sa
,tmp
,h
;
inline int read(){
int x=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x;
}
inline bool jud(int x){
int res=0;
for(int i=1;i<=n;++i){
if(h[i]>=x){if(++res==m-1) return true;}
else res=0;
//如果中断了,那目前满足条件的这个字符串就不会再出现了,所以清零,重新计数
}return (res>=m-1);
}
int main(){
//	freopen("a.in","r",stdin);
n=read();m=read();
for(int i=1;i<=n;++i) a[i]=aa[i]=read();
std::sort(aa+1,aa+n+1);
for(int i=1;i<=n;++i) a[i]=std::lower_bound(aa+1,aa+n+1,a[i])-aa;
memset(count,0,sizeof(count));
memset(rank,0,sizeof(rank));
for(int i=1;i<=n;++i) count[a[i]]=1;
for(int i=1;i<=n;++i) count[i]+=count[i-1];
for(int i=n;i>=1;--i) rank[i]=count[a[i]];
int k=0;
for(int p=1;k!=n;p<<=1){
memset(count,0,sizeof(count));
for(int i=1;i<=n;++i) count[rank[i+p]]++;
for(int i=1;i<=n;++i) count[i]+=count[i-1];
for(int i=n;i>=1;--i) tmp[count[rank[i+p]]--]=i;
memset(count,0,sizeof(count));
for(int i=1;i<=n;++i) count[rank[tmp[i]]]++;
for(int i=1;i<=n;++i) count[i]+=count[i-1];
for(int i=n;i>=1;--i) sa[count[rank[tmp[i]]]--]=tmp[i];
memcpy(rank1,rank,sizeof(rank1));
rank[sa[1]]=k=1;
for(int i=2;i<=n;++i){
if(rank1[sa[i]]!=rank1[sa[i-1]]||rank1[sa[i]+p]!=rank1[sa[i-1]+p]) ++k;
rank[sa[i]]=k;
}
}k=0;
for(int i=1;i<=n;++i){
if(rank[i]==1){h[1]=0;continue;}
if(i==1||h[rank[i-1]]<=1) k=0;
if(k)--k;
while(a[i+k]==a[sa[rank[i]-1]+k]) ++k;
h[rank[i]]=k;
}
int l=1,r=n,ans=0;
while(l<=r){
int mid=(l+r)>>1;
if(jud(mid)){ans=mid;l=mid+1;}
else r=mid-1;
}
printf("%d",ans);
return 0;
}