出现k次以上重复的最长子串 poj 3261

先二分答案，然后分组。只要某一组包含的后缀数量大于等于k，表示有解

//出现k次以上重复的最长子串#include <iostream>#include <string>#include <cmath>#include <map>using namespace std;const int N=30000;int wa[N],wb[N],wv[N],wsum[N];int height[N],sa[N],rank[N];int n,k;int str[N];int ans;int cmp(int *r,int a,int b,int l){return r[a]==r[b] && r[a+l]==r[b+l];}void da(int *r,int *sa,int n,int m){int i,j,p,*x=wa,*y=wb,*t;for(i=0;i<m;i++)wsum[i]=0;for(i=0;i<n;i++)wsum[x[i]=r[i]]++;for(i=1;i<m;i++)wsum[i]+=wsum[i-1];for(i=n-1;i>=0;i--)sa[--wsum[x[i]]]=i;for(j=1,p=1;p<n;j*=2,m=p){for(p=0,i=n-j;i<n;i++)y[p++]=i;for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;for(i=0;i<n;i++)wv[i]=x[y[i]];for(i=0;i<m;i++)wsum[i]=0;for(i=0;i<n;i++)wsum[wv[i]]++;for(i=1;i<m;i++)wsum[i]+=wsum[i-1];for(i=n-1;i>=0;i--)sa[--wsum[wv[i]]]=y[i];for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;}}void calheight(int *r,int *sa,int n){int i,j,k=0;for(i=0;i<=n;i++)rank[sa[i]]=i;for(i=0;i<n;height[rank[i++]]=k)for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);}bool check(int x){int i,cnt=1;for(i=2;i<=n;i++){if(height[i]>=x){cnt++;if(cnt>=k)return 1;}elsecnt=1;}if(cnt>=k)return 1;return 0;}int main(){int i,j,M,l,r,mid;while(~scanf("%d%d",&n,&k)){M=0;for(i=0;i<n;i++){scanf("%d",&str[i]);str[i]++;if(M<str[i])M=str[i];}M++;str[n]=0;da(str,sa,n+1,M);calheight(str,sa,n);l=0;r=n-1;while(l<=r){mid=(l+r)>>1;if(check(mid)){ans=mid;l=mid+1;}else{r=mid-1;}}printf("%d\n",ans);}return 0;}