poj3261 （二分+后缀数组）可重叠的至少为k次最长重复字串

#include<stdio.h>#include<string.h>#define max 1000100int rank[max],wa[max],wb[max],wv[max],wn[max];int height[max],sa[max],r[max];int N,K;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++)wn[i]=0;    for(i=0; i<n; i++)wn[x[i]=r[i]]++;    for(i=1; i<m; i++)wn[i]+=wn[i-1];    for(i=n-1; i>=0; i--)sa[--wn[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++)wn[i]=0;        for(i=0; i<n; i++)wn[wv[i]]++;        for(i=1; i<m; i++)wn[i]+=wn[i-1];        for(i=n-1; i>=0; i--)sa[--wn[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++;    }    return;}void calheight(int* r,int* sa,int n){    int i,j,k=0;    for(i=1; 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++);    return;}int main(){    int i,j,sum,flag,ans,m;    while(scanf("%d%d",&N,&K)!=EOF)    {        m=0;        sum=0;        ans=0;        for(i=0; i<N; i++)        {            scanf("%d",&r[i]);            if(m<r[i])                m=r[i];        }        r[N]=0;        da(r,sa,N+1,m+1);        calheight(r,sa,N);        int s=0,e=N;        while(e>=s)        {            flag=0;            int mid=(e+s)>>1;            for(i=1; i<=N; i++)            //height[]存的是排名相邻的两个后缀的最长公共前缀                if(height[i]>=mid)                    sum++;                else                {                    if(sum+1>=K)                    {                        ans=mid;                        flag=1;                    }                    sum=0;                }            if(sum+1>=K)            {                ans=mid;                flag=1;            }            sum=0;            if(flag)                s=mid+1;            else                e=mid-1;        }        printf("%d\n",ans);    }    return 0;}